If you find a mistake, omission, etc., please let me know by e-mail.

The orange ball marks our current location in the course, and the current problem set.

Remarks and corrections for the D&F “Preliminaries” [D&F = Dummit and Foote’s Abstract Algebra = course textbook]:
• The letters $\mathbb Z$, $\mathbb Q$, etc. are “blackboard bold”, originating as a typewritten and then handwritten proxy for the printed bold $\mathbf Z$, $\mathbf Q$, etc. for the integers, rationals, etc. Nowadays they are also sometimes seen in typeset mathematics (as in D&F), but you’re welcome to use the traditional ordinary blackboard $\mathbf Z$, $\mathbf Q$, etc. as well if you write up your homework in LaTeX or the like. Just don’t use ordinary $Z$ for either $\mathbb Z$ or $\mathbf Z$ (and likewise Q, R, etc.), because that letter might be needed for some other set or variable or the like.
» Possibly the rationals are denoted by $\mathbf Q$ because “$\mathbf R$” was already in use for the reals and the letter Q suggests “quotients” (or a cognate word in some other European language).
• D&F write about the notation $f^{-1}(C)$ for the preimage of $C$ (when $C$ is a subset of the “codomain” $B$ of some function $f : A \to B$): “Note that $f^{-1}$ is not in general a function [from $B$ to $A$] and that the fibers of $f$ generally contain many elements since there may be many elements of $A$ mapping to the element $b$ [of $B$].” Maybe better to write that in general the fibers may contain many elements of $A$; a fiber may also contain few elements, or none at all.
• The first of D&F’s ten “properties of the integers” (Section 0.2) is the well ordering of ${\mathbf Z}^+$, the set of positive integers — not of ${\mathbf Z}$ which is not well-ordered (do you see why?). Since a minimal element of $A \subseteq{\mathbf Z}^+$, if it exists, is unique, we will usually write and say “the minimal element”.
• Note that Section 0.1 concludes “Finally, we shall assume the reader is familiar with proofs by induction”; possibly you’re not as familiar with well ordering, but that turns out to be equivalent to proof by induction. Thus to derive well ordering from induction, we prove the equivalent statement that if $A \subseteq {\mathbf Z}^+$ does not have a minimal element then $A$ is empty, which in turn means that $n \notin A$ for all $n \in {\mathbf Z}^+$. We prove by induction that for each $n \in {\mathbf Z}^+$ none of the positive integers $m \leq n$ is in $A$.
$\circ$ Base case, $n=1$: if $1 \in A$ then $1$ is minimal, contradiction.
$\circ$ Induction step: if $A$ contains no $m \leq n$ then $A$ cannot contain $n+1$ because then $n+1$ would be minimal. Hence $A$ does not contain $n+1$ either, and thus contains no $m \leq n+1$. This completes the induction step and the proof. ☐
Conversely, to prove induction from well-ordering: if we want to prove “for all $n \in {\mathbf Z}^+$, <statement $S(n)$>”, and have checked $S(1)$ and proved $S(n) \Rightarrow S(n+1)$, then the set $A := \{ n \in {\mathbf Z}^+ : S(n) \text{ is false}\}$ has no minimal element, so by well-ordering $A = \emptyset$, which means that indeed $S(n)$ is true for all $n$. ☐
• Besides “$e$”, other common notations for the identity element of a group $G$ are “${\rm id}$” (LaTeX: {\mathrm id}) or simply “1” (as D&F notes). If there is risk of confusion — e.g. between the group identity 1 and the number 1, or between the identity elements of different groups — we can specify $e_G$, $1_G$, etc. (The letter e might be from German Eins “one” or Einheit “unit, the number 1”.)
• Note that the proof of Proposition 1, part 2 uses only $a \star b = e = c \star a$, not the full hypothesis (iii) that $b$ and $c$ are two-sided inverses of $a$. Thus it is enough to require that each $a \in G$ has both a left and a right inverse; it then automatically follows that these inverses are equal to each other and to any other group element $b$ satisfying $a \star b = e$ or $b \star a = e$. [This is also our first example of the crucial role of the associativity axiom $(a \star b) \star c = a \star (b \star c)$.] Likewise for Part 1 the analysis $f = e \star f = e$ requires only that $e,f$ satisfy $e \star g = g$ and $g \star f = g$ for all $g$.

First problem set: Introduction to groups

NO CLASS SEPTEMBER 7

September 12: Basic group properties and examples (finish D&F 1.1); dihedral groups (D&F 1.2); if time: start on symmetric groups (D&F 1.3)

The initial examples of groups in D&F 1.1 (page 17) are familiar but atypical in that all are commutative. Nor can our two constructions of new groups from old (subgroups and direct products, see problems 4, 5 on the first problem set) produce non-commutative groups from commutative ones. The next sections of Chapter 1 introduce examples of non-commutative groups coming from geometry (dihedral groups), combinatorics (symmetric groups), and algebra (matrix and quaternion groups). We shall see later that there are actually considerable overlaps among these approaches; for example every dihedral group — and indeed any finite group — can be found as a subgroup both of a symmetric group and a matrix group.

Further remarks on D&F 1.1 and 1.2
• About part (5) of the first Proposition of D&F 1.1 (the “generalized associative law”): Already in part (4) we in effect used the $n=4$ case, for which there are five possible ways to compute $a_1 \star a_2 \star a_3 \star a_4$: $$ 1(2(34)), 1((23)4), (12)(34), (1(23))4, ((12)3)4. $$ How many choices are there for $n=5$ and beyond? For $n=1,2,3,4,\ldots$, the sequence of counts starts $1, 1, 2, 5, 14, 42, 132, 429, \ldots$; for general $n$, the number of choices is the $(n-1)$-st Catalan number, given by the equivalent formulas $(2n-2)! \, / \, ((n-1)!n!) = {2n-2 \choose n-1} - {2n-2 \choose n}$. (For example, $2 = 4!/(2!3!) = 6-4$, and $5 = 6!/(3!4!) = 20 - 15$.) There are many proofs known — the Wikipedia page gives five — though all take some work. The Catalan numbers are remarkably ubiquitous in and near combinatorics; this excerpt from Stanley’s Enumerative Combinatorics is an excellent and still far from exhaustive sample.
• For $n=1,2,3$ the “Latin square” property of the multiplication table of the group $G$ of $n$ elements (see the Problem 1 on the first problem set) suffices to determine all such $G$. But this soon becomes impractical once $n$ gets at all large; according to the Wikipedia page on Latin squares, all $n \times n$ Latin squares are equivalent for $n=1,2,3$, but there are $4$ possibilities for $n=4$, then $56$ for $n=5$ and the counts quickly explode from there on (and explode much faster than Catalan numbers). For comparison, we shall soon see that there are only $2$ possible groups of size $4$, and (even easier) only one of size $5$; for further counts, see Sequence #1(!) on the OEIS = Online Encyclopedia of Integer Sequences. [The Catalan numbers are Sequence #108.]
• Warning on notation for the dihedral groups: In many sources (including, last I checked, Wikipedia’s “Dihedral group” entry) the group of symmetries of a regular $n$-gon is called $D_n$, not $D_{2n}$. In Math 122 we shall use our textbook’s notation: $D_{2n}$ is the dihedral group with $2n$ elements.
• We can avoid having to keep track of edge cases involving vertices $1,2,n$ etc. by labeling the vertices by integers mod $n$. Then the rotation $r$ takes $x$ to $x+1$ (even for $x=n$), and the reflection $s$ takes $x$ to $2-x \bmod n$; the dihedral group consists of all permutations of the form $r^i(x) = x+i \bmod n$ and $s r^i(x) = 2-(x+i) = (2-i) - x \bmod n$ (or more convenientsly $r^i$ and $s r^{2-i}$, which takes $x$ to $i-x$) for $i \in {\bf Z} / n {\bf Z}$.
• Besides the finite dihedral groups $D_{2n}$ there is also the “infinite dihedral group $D_\infty$”, consisting of symmetries of $\bf Z$ (imagine the vertices of a regular $\infty$-gon). Like $D_{2n}$, this group $D_{\infty}$ consists of the powers $r^j$ of the symmetry $r$ that takes each $i \in {\bf Z}$ to $i+1$, together with elements $s r^j$ where $s$ takes each $i \in {\bf Z}$ to $-i$. There is a presentation by generators and relations, like formula (1.1) (D&F p.26) but dropping the relation $r^n = 1$ because $r$ is of infinite order: $$ D_\infty = \langle r, s \mid s^2 = 1, r s = s r^{-1} \rangle. $$ [BTW please use LaTeX commands \langle and \rangle, not < and >, to get the left and right angle brackets $\langle$ and $\rangle$: not only do < and > look wrong, but they also combine wrong: “$D_\infty = < r, s \mid s^2 = 1, r s = s r^{-1} >$”. Also, I got the spaces around the “|” by writing \mid instead of |.]
» Why “dihedral”? You’ve probably already seen the words tetrahedron, octahedron, etc. for solids with $4, 8, \ldots$ faces (so a cube is a regular hexahedron). Their groups of rigid symmetries are known as the tetrahedral, octahedral, etc. groups; these groups are the topics of Exercises 9–13 (p.28) for D&F 1.2, and appear a few further times in the text — see “Platonic solids, symmetries of” in the Index, page 928. Well, a “dihedron” would be a “solid” with only two faces, and then those sides must be just one polygon considered as the top and bottom of a zero-width prism; in a “regular dihedron”, these faces are two congruent regular $n$-gons for some $n$ — imagine a coin of zero width whose shape is not the usual circle but a regular $n$-gon. The group of rigid symmetries of such an object would be $D_{2n}$, with $r^j$ rotating the coin and $s r^j$ flipping it to switch its obverse and reverse (a.k.a. “heads” and “tails”) sides.
• Warning about generators and relations: the groups $D_{2n}$ are somewhat atypical in being so easily accessible via a generators-and-relations presentation $G = \langle S \mid R_1, \ldots, R_m \rangle$. In many cases it is not easy to understand a group given in this way, nor to decide for some $G$ and $S$ whether the relations $R_1,\ldots,R_m$ suffice to recover $G.$ In fact it is a theorem (Wikipedia cites papers by Novikov and Boone published 1955 and 1958) that there is no algorithm that can decide, given $S$, $R_1,\ldots,R_m$, and some “word” $w$ (a finite product of elements of $S$ and their inverses), whether $w$ is the identity element of $\langle S \mid R_1, \ldots, R_m\rangle$. In fact one can even specify $S$ and the $R_i$ in advance! See Wikipedia’s article “Word problem for groups” for an explicit choice with $\#(S)=10$ and $m=27$.
𝄞 Stick around for a few minutes after 4:15 for why some music theorists care about the dihedral group D₂₄.

September 14: Symmetric groups (D&F 1.3), matrix groups (1.4), and the quaternion group $Q_8$ (1.5)

Remarks on D&F 1.3
» $\Omega$ is the capital form of Greek letter omega (which is why the symbol is also used for the unit of electrical resistance named for Georg Ohm). As you might expect, the (La)TeX code is \Omega. The name literally means “big o”, distinguished from omicron = “little o”.
• Some familiar examples of permutations in the Real World™ include shuffling a deck of cards, scrambling a Rubik’s cube, and sorting a list. (For example, the fact that a “bubble sort” terminates implies that $S_n$ is generated by the $n-1$ “simple transpositions” $(1\;2)$, $(2\;3)$, … $(n-1\;n)$ of consecutive numbers in $\{1,2,3,\ldots,n\}$.) We have also seen that for any element $g$ of a group $G$ (finite or not), the map taking $x \in G$ to $gx$ is a permutation, as is the map taking $x$ to $xg$; thus both maps can be regarded as elements of the symmetric group $S_G$. What are the inverses of these permutations of $G$?

Remarks on D&F 1.4 and 1.5
• To show that if $p$ is prime then ${\bf F}_p$ is a field we need the fact that all nonzero elements of ${\bf F}_p$ have multiplicative inverses. This in turn comes down to item (7) in the “Properties of the Integers” in Section 0.2 of the Preliminaries (bottom of page 5): if $a \bmod p$ is nonzero then, taking $b=p$, we have $\gcd(a,b)=1$, so we can find integers $x$ and $y$ such that $ax + by = 1$; since $b=p$ this makes $ax \equiv 1 \bmod p,$ and we have found a multiplicative inverse $x$ of $a \bmod p.$
• In 1.4, for “… and whose determinant is nonzero” (just before the displayed equation for $GL_n(F)$) you can substitute “that are invertible”, which is equivalent and conceptually simpler.
• You might remember that the identification of matrices with linear transformations gives the conceptual explanation for the associativity of the matrix product. It also lets us regard $GL_n(F)$ as a subgroup of the group $G_{F^n}$ of permutations of the vector space $F^n$. (In general it’s a very small subgroup of $S_{F^n}$; for starters $0$ cannot move, but even in $S_{F^n - \{0\}}$ most permutations are not linear — the counts coincide only for $F = {\bf Z}/2{\bf Z}$ and $n=1,2$.)
• In the next problem set we shall exhibit a finite subgroup of $GL_2({\bf C})$ with the same structure as $Q_8,$ which lets us deduce that $Q_8$ is a group without directly checking each of the $7^3$ cases of the associative law that do not involve the identity.

Second problem set: Examples of groups
corrected 15.ix.2023 [Amy Feng]: the second part of Problem 3 concerns a regular octahedron, not a regular tetrahedron.
corrected 16.ix.2023 [Michael Moorman]: in problem 2, the group $G$ should have been (and now is) $D_{2n}$.

September 19 and 21: Homomorphisms and isomorphisms between groups (D&F 1.6); group actions (D&F 1.7)

Remarks on D&F 1.6 and 1.7
• You might expect the definition of a group homomorphism $\varphi$ from $G$ to $H$ (bottom of page 36) to include the requirement that $\varphi$ respect identities and inverses, i.e., that $\varphi(1_G) = 1_H$ and $\varphi(x^{-1})=(\varphi(x))^{-1}$ for all $x \in G$ (note that in the second equation the left-hand side is the inverse in $G$, and the right-hand side is the inverse in $H$). This in fact follows from the given definition; we’ll show $\varphi(1_G) = 1_H$ in class, and leave $\varphi(x^{-1})=(\varphi(x))^{-1}$ as an exercise (part of Exercise 1).
» The Greek letter $\phi$ (capital form: $\Phi$) is called “phi” (LaTeX: \phi or \varphi, \Phi); its name can be pronounced either “fie” or “fee”, either of which is correct — just not “foe” or “fum”.
• As with linear transformations, if $\varphi: G \to H$ and $\psi: H \to K$ are group homomorphisms then so is their composition $\psi \circ \varphi: G \to K$. [Here $\psi$ is the Greek letter psi, usually pronounced “sigh” but sometimes “psigh” or “psee”; its capital form is $\Psi$, and naturally the LaTeX commands are \psi, \Psi.]
• The “kernel” of an action of $G$ on $B$ (end of example 1 on page 43) is indeed the kernel of the associated homomorphism $G \to S_B$ (see Exercise 14 for the previous section, page 40).
• The regular left action of a group on itself (last example on page 44) already gives us Cayley’s theorem, which appears later in D&F as Corollary 4 on page 120: any group $G$ is isomorphic with a subgroup of $S_G$. This may seem to hardly merit the name of a “theorem”; but it is historically important (resolving a dispute on whether group theory should concern just subgroups of $S_\Omega$ or any $(G,\star)$ satisfying the group axioms: by Cayley the two notions are equivalent), and for finite $G$ this regular representation of $G$ also figures in the theory of linear representations (homomorphisms from $G$ to ${\rm GL}_n({\bf C})$, a.k.a. actions of $G$ on finite-dimensional complex vector spaces).

Third problem set: Group homomorphisms [including isomorphisms] and actions
corrected 23.ix.2023 [Marcello Laurel]: in problem 4 the formula defining the kernel should have (and now has) “$\varphi(g) = 1_H$”, not “$\varphi(G) = 1_H$”. The textbook has the correct formula (p.40, Exercise #14 for D&F 1.6).

September 26: More on subgroups: centralizers and normalizers, stabilizers and kernels (D&F 2.1 and 2.2)

Remarks on D&F 2.1 and 2.2
• D&F 2.1 is largely a review of material we already developed in class and/or in the assigned problems. The only new elements are the notations $H \leq G$ and $H \lt G$ for “$H$ is a subgroup of $G$” and “$H$ is a proper subgroup of $G$”, and the “Subgroup Criterion” (Proposition 1 on page 47; in practice it is about as easy to check directly that $H$ contains $1_G$ and is closed under both $xy$ and $x^{-1}$). Happily the notations for subgroups are consistent with our expectation that if $H \leq G$ and $K \leq H$ then $K \leq G$ (and $K \lt G$ provided at least one of the inclusions $H \leq G$ and $K \leq H$ is proper).
• Example (2) on page 47: except if $G$ is the trivial group, in which case it has only one subgroup… You can already work out which nontrivial groups $G$ have no subgroups other than $\{1\}$ and $G$ (the answer is suggested by Lagrange’s theorem [D&F Exercise 1.7#19 on page 45, #6 in the latest problem set], though Lagrange is not really needed for this purpose).
• The same notation $C_G$ is used for the centralizers of an element of $G$ and of a subset of $G$ (so $C_G(a)$ is also $C_G(\{a\})$ ).
• The term “centralizer” is used because $C_G(a)$ is the largest subgroup of $G$ whose center contains $a$ (check that $C_G(a)$ does always contain $a$!). Likewise for $C_G(A)$. There is a similar rationale for “normalizer”, but we have not touched on normal subgroups yet.
• If $G$ acts on some set $\Sigma$ then $G$ also acts on the power set ${\cal P}(\Sigma)$ (a.k.a. $2^\Sigma$) by $g \cdot A := \{ g \cdot a \mid a \in A \}$. If $\Sigma = G$ with $G$ acting on itself by conjugation then the stabilizer of $A$ is the normalizer $N_G(A)$.
• See the Exercises for D&F 2.1 for further examples of subgroups and subgroup constructions.

September 28: The structure of cyclic (sub)groups (D&F 2.3)

Remarks on D&F 2.3
• As the textbook more-or-less notes, Proposition 2 is basically a review of our earlier description of the order of an element. (There is no Proposition 1 in this section; D&F numbers Propositions and the like by chapter, and this chapter’s Proposition 1 is the “Subgroup Criterion” of 2.1 . Note also that the next numbered result after Proposition 3 is Theorem 4 — and there is no Theorem $n$ for $n \in \{1,2,3\}$ in this chapter.)
• In Proposition 3, “$(m,n)$” is not an ordered pair but a common abbreviation for $\gcd(m,n)$ (gcd = greatest common divisor); likewise $(n,a)$ in Proposition 5, $(a,n)$ in Proposition 6, and $(n,b)$ in Theorem 7. Alternatively: by this point we already know that if $x^n = 1$ then $|x|$ is (finite and) a factor of $n$; thus if $x^m = x^n = 1$ then $|x|$ is a factor of both $m$ and $n$, and thus of $\gcd(m,n)$. The approach via the Euclidean Algorithm prefigures our later development of principal ideas in commutative rings.
• In Theorem 4, the point of the claim that $\varphi$ is well-defined is that there can be more than one way to write an element of $\langle x \rangle$ as $x^k\!$, but all of them yield the same element $y^k$ of $\langle y \rangle$.
• It might help to visualize Proposition 5 (page 57) as follows. The group element $x^a$ goes $a/n$ of the way around the cycle. Then $(x^a)^m$ goes $ma/n$ of the way around, and we want to make that an integer. Reduce $a/n$ to lowest terms as $a'/n'$. Then $ma/n = ma'/n'$, and since $a'/n'$ is in lowest terms its $m$th multiple is an integer iff $m$ is a multiple of $n$. Now observe that $n' = \frac{n}{\gcd(n,a)}$.
» Merriam–Webster knows about iff; note the pronunciations! The Wiktionary entry for iff reports that several other languages calqued the abbreviation: Spanish sii, Finnish joss, Swedish omm, Danish and Norwegian hviss, and even Hebrew IMM with an incongruous doubled final-Mem.
• Note that the $Z$ in the notation $Z_n$ for a cyclic group of order $n$ is an ordinary italic $Z$, not the Z or ℤ used for the integers. This $Z$ comes from a different German word, zyklisch (“cyclic”). The notation $C_n$ is also used (where I guess the $C$ is not directly from our “cyclic” but from French “cyclique”).

Fourth problem set: Subgroups
clarification 3.x.2023: yes, in 2(ii) we retain the requirement $\forall n: H_n \leq H_{n+1}$ of 2(i).

October 3: Subgroups generated by subsets of $G$ (D&F 2.4)
(and the conclusion of 2.3)

Remarks on D&F 2.4
• As with centralizers, we use the notation $\langle \cdots \rangle$ both for the subgroup $\langle A \rangle$ generated by a subset $A$ of a group and for the subgroup $\langle a_1, a_2, \ldots \rangle$ generated by some elements of the group. For example, $\langle a_1, a_2, \ldots, a_n \rangle$ means the same as $\langle \{ a_1, a_2, \ldots, a_n \} \rangle$.
• In particular, if $n=1$ we recover our notation $\langle x \rangle$ for the cyclic subgroup generated by a single group element $x$.
• At the end of D&F 2.3 we saw that a subgroup of a cyclic group is again cyclic. Since “cyclic” means the same as “one-generator“, one might be tempted to guess that likewise a subgroup of an $n$-generator group can be generated by at most $n$ elements of the subgroup. Warning: this is false in general! For each integer $N$ there are finite two-generator groups $G$ with a subgroup $H$ that cannot be generated by fewer than $N$ elements of $H$. For instance, the textbook asserts (but does not yet prove) that every symmetric group $S_n$ ($n \geq 2$) can be generated by two permutations, such as the simple transposition (1 2) and the $n$-cycle (1 2 $\ldots$ $n$); once $n \geq 2N$ this group $S_n$ has a subgroup that requires $N$ generators, namely the subgroup generated by $N$ disjoint simple transpositions. [Since this subgroup is abelian, and every element has order 1 or 2, any $k$ elements generate a subgroup of size at most $2^k.$] In the next problem set we also give an example where $G$ is infinite and $H$ has no finite generating set at all.
• We shall probably see later in Math 122 that if $G = \langle a_1, a_2, \ldots \rangle$ is abelian then any subgroup of $G$ can be generated by at most $n$ generators. This generalizes the case $n=1$ from the end of D&F 2.3 (where $G$ is automatically abelian).

October 5: Normal subgroups and quotient groups (D&F 3.1)

The $\triangleleft$ symbol is available in LaTeX as \triangleleft . The amssymb package also provides a \trianglelefteq symbol that can be used to specify “is a normal group which need not be a proper subgroup”.

Remarks on D&F 3.1
• We can now explain the term “normalizer” as an analogue of “centralizer”. The centralizer $C_G(A)$ of some $A \subseteq G$ is the largest subgroup $G' \subseteq G$ whose center contains $A$; in particular $C_G(A) = G$ if and only if $A \in Z(G),$ so we might say $A$ is central in $G$. Likewise, if $A$ is a subgroup of $G$ then its normalizer $C_G(A)$ is the largest subgroup $G' \subseteq G$ in which $A$ is a normal subgroup, and in particular $N_G(A) = G$ if and only if $A$ is normal in $G$. (Yes, this analogy is not perfect as it stands, because $A$ did not have to be a subgroup for our description of $C_G(A)$; but we know that $C_G(A) = C_G(\langle A \rangle)$, so we do not really lose any generality by assuming that this $A$ is also a subgroup.)
» The word “normal” is known for having many unrelated technical uses in mathematics. The four senses currently included in the Merriam–Webster entry (and quoted in the fifth problem set) are only a fraction of the list in Wikipedia’s dismabiguation page (where ours appears last because the list is alphabetized), even accounting for the fact that some entries in this list are closely related (such as normal matrix and normal operator).
• Warning: unlike $\subset$ and $\leq$, the relation $\triangleleft$ is not transitive: there can be groups $G'' \subset G' \subset G$ such that $G'' \triangleleft G'$ and $G' \triangleleft G$ but $G''$ is not a normal subgroup of $G$. This already happens for $G = D_8$; can you find appropriate $G'$ and $G''$ in this case? (See D&F 3.2, item (3) on page 91.)

kifth problem set: Subgroups generated by subsets; normal subgroups
corrected 7.x.2023 [Wittmann Goh]: in 2(ii), matrices of the form $({1 \; b \atop 0 \; 1})$ with $b \in {\bf R}$ [not “$a \in {\bf R}$”]

October 10: Cosets, continued (D&F 3.2)

Remarks on D&F 3.2
• The opening sentence of this section feels misleading: it says “In this section we continue the study of quotient groups” but most of the discussion is about cosets of subgroups that are typically not normal and thus do not correspond to a quotient group.
• Lagrange’s Theorem (Thm.8 on page 89–90) and its proof already appeared in the third problem set (problems 5 and 6), though without the context of cosets.
• Other notations for the index of a subgroup $H$ of a group $G$ are $[G:H]$ and (per Wikipedia) $(G:H)$. I’ll try to stick to D&F’s $|G:H|$ but you might on occasion see $[G:H]$ which is the notation I'm used to.
• This chapter of the text does not seem to state explicitly that there is a bijection between left cosets $gH$ and right cosets $Hg'$, so in particular $(G:H)$ is also the number of right cosets even if we do not assume that $|G| \lt \infty.$ Indeed the image of $gH$ under the inverse map $x \mapsto x^{-1}$ is $H^{-1} g^{-1} = H g^{-1}$. (This argument is given in D&F 3.2, Exercise 12.) The important special case $(G:H) = 2$ of this is used in the proof that subgroups of index 2 are always normal (see page 91, Example 2).
• A further corollary of Lagrange’s Theorem is that if $G$ is a group of prime order then the only subgroups of $G$ are $\{1\}$ and $G$ itself. Conversely, if $G$ is a finite group such that $|G|$ is not prime (nor equal to 1) then $G$ does have at least one other subgroup: let $x$ be any nontrivial element; then $\langle x \rangle$ is a subgroup other than $\{1\}$, and if $\langle x \rangle = G$ then $|x| = |G|$ — but $|G|$ is not prime so we can write $|G| = mn$ for some integers $m,n\gt1$ and then use the subgroup $\langle x^m \rangle.$ [I called this “Proposition $10\frac12$” in class.]
• Proposition 14 (page 94) gives another example of the observation that practically anything that can be proved with the “Subgroup Criterion” (D&F 2.1) can be done about as easily without it. Thus if $H,K$ are subgroups of $G$ such that $HK = KH$ then:
$\circ$ $1 \in HK$ because [why?];
$\circ$ $HK$ is closed under inverses because $(hk)^{-1} = k^{-1} h^{-1}$ and $H,K$ are closed under inverses: if $h\in H$ and $k \in K$ then $h^{-1}\in H$ and $k^{-1} \in K$ so $(hk)^{-1} \in KH = HK$; and
$\circ$ $HK$ is closed under products because $H$ and $K$ are: if $h,h' \in H$ and $k,k' \in K$ then $(hk)(h'k') = h(kh')k'$, and $kh' \in KH = HK$, so we can write $kh' = h'_1 k_1^{\phantom\prime}$ and deduce $(hk)(h'k') = h(h'_1 k_1^{\phantom\prime})k' = (h h'_1) (k_1^{\phantom\prime} k') \in HK.$

October 12: The isomorphism theorems, etc. (D&F 3.3)

Remarks on D&F 3.3
• Different authors use different numbers and names for what D&F call the First, Second, and Third Isomorphism Theorems. We’ll say little about the “Fourth Isomorphism Theorem“ (for which Wikipedia recognizes the names “lattice theorem“ and “correspondence theorem”).
• The connection between Corollary 17(2) and the rank-nullity theorem is not as immediate as D&F suggest. If $\varphi: V \to W$ is a homomorphism of vector spaces then there is a quotient vector space $V / \ker(\varphi)$ that is isomorphic with the image of $\varphi$. If $V$ is finite-dimensional then comparing dimensions yields rank-nullity. But this is not tantamount to $\left| V : \ker(\varphi) \right| = \left| \varphi(V) \right|$ unless $V,W$ are vector spaces over a finite field.
• Concerning the first full paragraph of page 100: an even simpler example uses the two groups ${\bf Z} / 4{\bf Z}$ and $({\bf Z} / 2{\bf Z}) \times ({\bf Z} / 2{\bf Z})$ of order 4: each of those $G$ has a (necessarily normal) subgroup $N$ with $|N| = |G/N| = 2,$ and all groups of order 2 are isomorphic.

Sixth problem set: Cosets, Cauchy, etc.
corrected 13.x.2023: the proof of Cauchy’s theorem counts as both #3 and #4 (h/t AnaMaria Perez); and in that problem the cardinality of $\cal S$ is $|G|^{p-1}$, not “$G^{p-1}$”, and the notation $\vec x$ for the $p$-tuple $(x_1,\ldots,x_p)$ is introduced together with that tuple.
corrected 14.x.2023: Likewise in 1(i) the cardinality of $gHg^{-1}$ is $|H|$, not $H$ (noted by Michael Moorman, Willa Fogelson, and, independently(?), Daniel Dickman).
corrected 15.x.2023: in part (ii) of problem “3–4” the trivial subgroup is $\{0\}$, not $\{1\},$ because we are using additive notation for ${\bf Z} / p {\bf Z}$ (noted by M. Moorman)
corrected 17.x.2023 ☹: in problem 5, $p$ is the index $|G:H|$, not $|H:G|$
corrected 18.x.2023: problems 5 and 6 are D&F 3.3 #3,8, not D&F 3.4 #5,8

October 17: The sign homomorphism $S_n \to \{1,-1\}$ and the alternating group $A_n$ (D&F 3.5)

There are many ways to construct the sign homomorphism, but none that feels well motivated. The difficulty can be understood as follows. Suppose there is a nontrivial homomorphism $h : S_n \to A$ for some abelian group $A$. We know that $S_n$ is generated by transpositions, which have order 2. Hence the image of h must consist of elements of $A$ of order 1 or 2. Moreover, all transpositions in $S_n$ are conjugate, from which it soon follows that they must have the same image in $A$, call it $z$. Since $h$ is assumed nontrivial, $z$ must have order 2. It follows that the image of any $\sigma \in S_n$ is $z^N$ where $\sigma$ is the product of $N$ transpositions. The rest of the properties of $h$ soon follow — assuming $h$ is well-defined, that is, assuming that $z^N$ does not depend on the representation of $\sigma$ as a product of transpositions. But this is not at all obvious: why can the same permutation not be realized both as a product of an even number and of an odd number of transpositions?

The usual way around this difficulty is to construct the map $h$ in some other way and then show that it is in fact a homomorphism, from which the other properties follow. Various ways to do this are known, none of which is ideal, though each of them tells us something more about $h$. We’ll take a route via the “inversion number” ${\rm inv}(\sigma)$ of a permutation $\sigma \in S_n$. An inversion of a permutation $\sigma$ is a pair $(i,j)$ of integers in $\{1,\ldots,n\}$ such that $i \lt j$ and $S(i) \gt S(j)$. We shall call a permutation even or odd according as its inversion number is even or odd. This agrees with the definition in D&F because ${\rm inv}(\sigma)$ is the number of negative factors $(\sigma(i) - \sigma(j)) \, / \, (i-j).$ (*) The inversion number is also the number of steps it takes to bubble-sort $\sigma$. We shall show that the inversion number has the following properties:

• ${\rm inv}(\sigma) \geq 0$, and ${\rm inv}(\sigma) = 0$ iff $\sigma$ is the identity permutation.
• Composing $\sigma$ (from either side) by a consecutive simple transposition — that is, a simple transposition $(i\;j)$ with $j = i+1$ — changes ${\rm inv}(\sigma)$ by adding $+1$ or $-1$.
• As long as $\sigma$ is not the identity permutation there is at least one choice of $i$ for which ${\rm inv}( (i{\phantom{0,}}i+1) \, \sigma) = {\rm inv}(\sigma) - 1$.

(*) D&F define the sign by $\epsilon(\sigma) = \prod_{i\lt j} (x_i - x_j) = \prod_{i\lt j} (x_{\sigma(i)} - x_{\sigma(j)})$. Choose $x_1,\ldots,x_n = 1,\ldots,n$ (i.e. $x_i = i$ for each $i$), and solve for $\epsilon(\sigma)$ to get $\epsilon(\sigma) = \prod_{i\lt j} (\sigma(i) - \sigma(j)) \, / \, (i-j)$. So, since $\epsilon(\sigma)$ is either $1$ or $-1$, we can find $\epsilon(\sigma)$ by counting how many factors $(\sigma(i) - \sigma(j)) \, / \, (i-j)$ are negative: $\epsilon(\sigma)$ is $+1$ if the number is even, and $-1$ if odd. But $(\sigma(i) - \sigma(j)) \, / \, (i-j) \lt 0$ exactly when $(i,j)$ is an inversion of $\sigma$. Thus $\epsilon(\sigma) = (-1)^{{\rm inv}(\sigma)}.$

One neat application is the unsolvability of Loyd’s 15-14 puzzle: can the 15 Puzzle be manipulated so as to return to its original state but with two tiles switched? Consider a position of the puzzle as a permutation of 16 objects: the 15 tiles and the hole. Note that the reachable configurations do not necessarily form a group, because we can only concatenate two move-sequences if their first one takes the hole to where it is needed for the start of the second sequence, so this subset of $S_{16}$ need not be closed under the group operation.* But we can still consider each move as a simple transposition switching a hole with a tile. When the hole returns to its initial place, it has made an even number of moves, because each move changes the hole’s color in a checkerboard coloring of the $4 \times 4$ square. Therefore the resulting permutation of the tiles and hole is even, and in particular cannot be a simple transposition, QED.

(*) On the other hand, the reachable configurations in which the hole has been brought back to the bottom right corner do allow for arbitrary concatenations, and so must form a subgroup of $S_{16}$. (Since they all stabilize the hole, they can then also be regarded as a subgroup of $S_{15}$.)

It is known that conversely all even permutations can be realized. The same Wikipedia page gives a reference to a 1879 paper by Johnson and Story proving this not just for the $4 \times 4$ puzzle but generally for any rectangular $m \times n$ puzzle as long as both sides $m$ and $n$ are at least 2. [Yes, that’s 1879, not 1978 or 1987; this was published in the second volume of the American Journal of Mathematics, which is now in its 145th year and volume.]

One sometimes sees 2-cycles called “simple transpositions” (not just “transpositions”, as in the Definition in D&F p.107), in contrast to order-2 permutations that are products of 2, 3, … disjoint 2-cycles, which are “double transpositions”, “triple transpositions” etc. Thus “simple” means “single” in this context, and also in a few other mathematical settings — you may have already encountered polynomials with simple roots, double roots, triple roots, etc. As it happens the Latin soruce simplex of “simple” already had “single” as one of its senses.

October 19: Group actions and permutation representations (D&F 4.1, 4.2)

Remarks on D&F 4.1 and 4.2
• Much of these two sections are review of material that we have already covered, which is why I’m slating them for a single lecture.
• The last sentence of Proposition 2 (page 114) is usually called the orbit-stabilizer theorem. D&F cannot call it that at this point because they have not yet defined “orbit” (which they do in the next page).
• In particular, this result together with Lagrange shows that if $G$ is finite then $|G|$ is the product of the sizes of the orbit and the stabilizer of $a$. The proof of Cauchy’s theorem already used the special case of this fact when $|G|$ is prime (so one of the two factors must be 1 and the other factor must equal $|G|$.)
• The proof in D&F of this result (that is, the end of the proof of Proposition 2) elides the key point that the map $b \mapsto g G_a$ is well-defined: if $b = g \cdot a = g' \cdot a$ then $g G_a$ and $g' G_a$ are the same coset. (To be sure the proof is straightforward.)
• Combining results of 4.1 and 4.2, together with 4.1 Exercise 1 (page 116), we see that transitive actions of a fixed group $G$ on an arbitrary set $A$ correspond to subgroups $H \subseteq G$ up to conjugation: from the action we recover $H$ as the stabilizer $G_a$ for any $a \in A$; and from $H$ we recover the action by letting $A$ be the set of left cosets of $H$.

Seventh problem set: The sign homomorphism and the alternating group; group actions cont’d
corrected 24.x.2023: in problem 4(i), both $m$ and $n$ must be at least $1$ (because $(S_1 : A_1) = 1$, not $2$ which is the index $(S_n : A_n)$ for all $n>1$). [Paula Leyes Carreño]; in the text following problem 4, I switched the exponents: should be “$2^{11} 3^7,$ not $2^{12} 3^8$” (there are 12 edge cubelets each with 2 possible orientations, and 8 edge cubelets each with 3 possible orientations), not “$2^7 3^{11},$ not $2^8 3^{12}$” as I wrote [Edvin Tewolde Berhane].

October 24: Groups acting on themselves by conjugation; the class equation of a finite group, and the simplicity of $A_5$ (D&F 4.3)

Remarks on D&F 4.3
• One could also write the class equation as $|G| = \sum_j |G : C_j(x_j)|$ where $x_j$ runs over representatives of all conjugacy classes, central or not. (That’s equivalent because each $x_j \in Z(G)$ contributes $1$ to the sum.) But the textbook’s choice of separating out the central terms seems to be universal.
• We already knew that a group of prime order $p$ is abelian (and in fact cyclic); Corollary 9 shows that all groups of order $p^2$ are abelian. This is no longer true for order $p^3$: we have already seen the example of $D_8$ and $Q_8$ (for $p=2$) and the groups of upper-triangular matrices with entries in ${\bf Z} / p {\bf Z}$ and 1’s on the diagonal (for any $p$).
• More generally if $G$ acts on any finite set $S$ then the orbit-stabilizer theorem gives $|S| = \sum_j |G : G_j|$ where $G_j$ is the stabilizer of a representative of the $j$-th orbit. A singleton orbit has stabilizer $G$, so we can also write $|S| = |S_0| + \sum_i |G : G_i|$ where $S_0$ is the fixed subset and $G_i$ is the stabilizer of a representative of the $i$-th orbit of size $> 1,$ see e.g. this page (which assumes $G$ is finite so that $|G : G_i|$ can be written as $|G| / |G_i|$).
• The number of partitions of $n$ (which is the number of conjugacy classes in $S_n$) grows quickly with $n$, faster than any power of $n$ though slower than $c^n$ for any $c \gt 1$. Believe it or not, the number of partitions of $n$ is asymptotic to $\exp (\pi \sqrt{2n/3}) \, / \, (4n\sqrt{3})$ as $n \to \infty$ (a famous theory of Hardy and Ramanujan). For more about the partition function, see the relevant OEIS and Wikipedia pages.
• The proof of the simplicity of $A_5$ works also for many other finite simple groups, including the two next-smallest groups, ${\rm PSL}_2({\bf Z}/7{\bf Z})$ (order 168, conjugacy group sizes $1, 21, 24, 24, 42, 56$) and $A_6$ (order $6!/2 = 360$, conjugacy classes of sizes $1, 40, 40, 45, 72, 72, 90$). But the computations become increasingly difficult (see the previous paragraph), and sometimes this approach fails, e.g. the nummber of $3$-cycles in $A_{68}$ is $68 \cdot 67 \cdot 66 \, / \, 3 = 100232,$ and $1 + 100232 = 3^2 \, 7 \; 37 \; 43$ is a factor of $68! / 2 = |A_{68}|$. This happens again for $A_n$ with $n = 149, 179, 247, 327, 347, 362, 369, \ldots$; there are probably infinitely many such $n$, though this might be quite hard to prove. To be sure the identity and 3-cycles in $A_n$ do not form a subgroup (they are very far from closed under multiplication), let alone a normal subgroup. We shall soon build on the simplicity of $A_5$ to prove that $A_n$ is simple also for all $n \gt 5$.

October 26: The group of automorphisms of a group $G$ (D&F 4.4)

Remarks on D&F 4.4
• In general an automorphism of any mathematical structure $M$ is a bijective isomorphism $M \to M$, i.e. a bijection that preserves the structure; and these bijections constitute a subgroup of $S_M$, called the automorphism group of $M$ and denoted by ${\rm Aut}(M).$ For example, an automorphism of a vector space $V$ is an invertible linear transformation from $V$ to itself (cf. Proposition 17, part 3); an automorphism of a topological space $X$ is a continuous bijection $f: X \to X$ such that $f^{-1}$ is also continuous; and an ”automoprhism“ of a set $\Omega$ with no further prescribed structure is just an arbitrary permutation of $\Omega$.
$\large\circ$ For more about the Petersen graph (which the topic of my extemporaneous digression on automorphism groups), see this Wikipedia page.
• The proof of Proposition 13 repeats steps we have done already, in some cases more circuitously than necessary, perhaps for didactic reasons — e.g. the fact that $\varphi_1$ is the identity map is immediate from its definition, and is the first thing we checked when verifying that conjugation indeed gives an action.
• Likewise we have already seen Proposition 14 (when noting that the class equation does not depend on the choice of representative $g_i$ of the conjugacy class); more generally $K \cong \varphi(K),$ and thus $|K| \cong |\varphi(K)|,$ for any automorphism $\varphi$ of $G$.
• Re Proposition 15 and the ensuing Definition: Exercise 1 for this section shows that the inner automorphism group ${\rm Inn}(G)$ is in fact a normal subgroup of ${\rm Aut}(G).$
• About the Definition on the top of page 135: The notation “$H$ char $G$” for “$H$ is a characteristic subgroup of $G$” is not standard (because “char” has been preempted by another mathematical notion of “characteristic”), and we shall not use it.
• About the Example on pages 135--136: one can more symmetrically require that $p \not\equiv 1 \bmod q$ and $q \not\equiv 1 \bmod p$; the former is automatic if we assume $p \leq q.$ We already know the existence of $x$ of order $q$ by Cauchy, so we do not need the argument at the top of page 136. We did not mention Corollary 5 in class, but the fact that $\langle x \rangle$ must be normal will also follow from the Sylow theorems that we shall prove in D&F 4.5.
$\large\circ$ Conversely, if $q \equiv 1 \bmod p$ then there does exist a non-abelian group $G$ of order $pq$. Let $F$ be the finite field ${\bf Z} / q {\bf Z},$ and let $P$ be a $p$-element subgroup of $F^\times,$ which exists by Cauchy. Then $G$ can be constructed as the subgroup of ${\rm GL}_2(F)$ consisting of matrices $\left({a\;b \atop 0\; 1}\right)$ such that $a \in P.$ (Once we know that $F^\times$ must be cyclic it will follow that it has a unique $p$-element subgroup, from which one can show using the techniques in D&F 4.4 that every non-abelian group of order $pq$ is isomorphic with this $G$.)
• Proposition 17, part 1: we already mentioned that $({\bf Z}/ p{\bf Z})^\times$ is cyclic for every prime $p$ (the existence of “primitive residues” modulo a prime), but did not prove it, and probably will not reach this result in Math 122. But all we shall use in this class is that $({\bf Z}/ p{\bf Z})^\times$ is an abelian group of size $p-1.$ Likewise we shall note but not prove parts 2, 4, and (probably) 5.

Eighth problem set: The action of $G$ on itself by conjugation: class equation, Aut$(G)$, etc.

October 31: The simplicity of $A_n$ for $n \gt 5$ (D&F 4.6); introduction to Sylow’s theorems (D&F 4.5)

Remarks on D&F 4.6
• This recursive/inductive proof of the simplicity of $A_n$ is a prototype of the proofs of simplicity of ${\rm PSL}_n(F)$ and other infinite families of simples groups. In those cases, too, there are sometimes special small cases that must be excluded.
• You might notice that the hypothesis $n \gt 5$ is not used until the end of the proof. In effect, then, we first prove that if $A_{n-1}$ is simple then so is $A_n$ unless $A_n$ contains a transitive normal subgroup $H$ with trivial stabilizer (which by orbit-stabilizer is equivalent to $|H| = n$), and then show that there is no such $H$ for $n$ large enough — $n \gt 5$ is sufficient for our purpose; we already know it is also true for $n=5,$ but we cannot use this to give an alternative proof of the simplicity of $A_5$ because $A_4$ is not simple! Indeed the normal subgroup $V$ of $A_4$ is transitive with simple stabilizer, as we now know it had to be because $A_3$ is simple.
• I’ll give the following version of the inductive step. Let $H$ be a normal subgroup and $h \in H.$ Prove as in D&F that if $h$ has a fixed point then $H = A_n.$ Then $H$ contains every conjugate $ghg^{-1}$ ($g \in G$), and thus every commutator $ghg^{-1}h^{-1}$. If $g$ is a 3-cycle then $ghg^{-1}h^{-1} = g (hg^{-1}h^{-1})$ is the product of a $3$-cycle and its conjugate, and thus moves at most 6 points. So at least if $n \gt 6$ it must be the identity. But $ghg^{-1}h^{-1} = 1$ if and only if $g$ is in the centralizer of $h$ — and the 3-cycles in $A_n$ generate $A_n$ so $h \in Z(A_n).$ Therefore $h=1.$ Since this is true for all $h \in H$ we conclude that $H = \{1\},$ QED. For $n=6$ we can take another commutator to show that once $H$ contains a double 3-cycle it contains a single 3-cycle, and thus all 3-cycles, and these generate $A_6$ so again we’re done.

Remarks on D&F 4.5
• If $P$ is a $p$-group of order $p^\alpha$ then $P$ contains a $p$-subgroup of order $p^\beta$ for each $\beta \leq \alpha.$ Thus Sylow’s first theorem also implies that if $G$ is any finite group whose order is a multiple of $p^\beta$ then $G$ contains a subgroup of order $p^\beta,$ even if $p^\beta$ is not the largest power of $p$ dividing $|G|$.
• Warnings: Suppose $G$ is a finite group whose order is a multiple of the prime $p.$ We already know from our proof of Cauchy’s theorem that the number of $p$-element subgroups of $G$ is congruent to $1 \bmod p.$ If $p$ is the largest power of $p$ dividing $G$ then all these groups are conjugate by Sylow 2; but this is not true in general. The group $S_4$ provides a counterexample with $(p,\alpha) = (2,3)$: There are two conjugacy classes of 2-element subgroups, generated by the 6 simple transpositions and the 2 double transpositions. Note that the first of these also gives an example where the number of conjugates is not $1 \bmod p.$ Nor is it necessarily true that the number of $p^\beta$-element subgroups is congruent to $1 \bmod p$ for every $\beta \in [1,\alpha].$ In view of Cauchy’s theorem and Sylow 3, a counterexample must have $\alpha \geq 3$ and $\beta \geq 2;$ and again $S_4$ provides a counterexample for $p=2$: the $4$-element subgroups are the Klein group $V$ and three cyclic subgroups. As expected from Sylow 2, each is contained in one of the Sylow subgroups of $S_4$, which are the three copies of $D_8$ in $S_4$.

Ninth problem set: Simplicity of $A_n$ ($n>5$); Sylow’s theorem(s)

November 2: Sylow’s theorems (D&F 4.5; we will not cover most of the applications to specific $|G|$)

Roadmap for the proofs of the Sylow theorems: Sylow 1 is proved by induction on the size of $G,$ using at each step Cauchy’s theorem in the case of abelian groups (D&F 3.4, Prop. 21) if $|Z(G)|$ is a multiple of $p,$ or the class equation if not. Once we have one Sylow subgroup $P$, we prove Sylow 2 and 3 by applying orbit-stabilizer and Proposition 19 to the action of $Q$ by conjugation on the set $\cal S$ of $G$-conjugates of $P$. Note that the proof of Proposition 19 uses Corollary 15 to Proposition 14 of D&F 3.2 (page 94), which in retrospect helps explain that apparent digression in the textbook.

Further remarks on D&F 4.5
• The next-to-last line of page 140 contains a rare lapse of mathematical writing: “By definition of $\cal S$, $G$, hence also $Q$…”. This is a kind of “garden-path sentence”: we expect “by definition of $\cal S$, $G$, and…” but it turns out to be “by definition of $\cal S$, the group $G,$ and hence also its subgroup $Q$” etc. I mention this not only to help you process this sentence but also to caution against this kind of thing in your own problem-set solutions and other mathematical writing — not just for your readers’ sake but also for your own when reviewing your solutions etc. A general rule of thumb that covers many such cases is to avoid having consecutive mathematical formulas separated by just a space or punctuation mark, except as part of an “and” or “or“ list (as in “a group isomorphic to $S_5$, $A_6$, or $S_6$”). An extreme example of what to avoid (which I learned from J.-P. Serre): “If $A$, $B$, $C$.” This is actually ambiguous: did the writer mean “if $A$ and $B$ then $C$” or “if $A$ then $B$ and $C$”?

November 7: Introduction to rings (D&F 7.1)

Remarks on D&F 7.1
• The text uses (e.g. in the description of arithmetic in the Hamilton quaternions), but does not seem to explicitly state or prove, the fact that the ring axioms imply the familiar rule for expanding $(a+b) \times (c+d).$ Using henceforth the convention of dropping the “$\times$” symbol, we have $$ (a+b) (c+d) = a (c+d) + b(c+d) = (ac + ad) + (bc + bd) = ac + ad + bc + bd. $$ Here we used the distributive law in the first two steps. This already puts the four terms in the order you may have learned in grade school by the mnemonic acronym FOIL (first, outer, inner, last). If we started instead with $(a+b) (c+d) = (a+b) c + (a+b) d$ we would end up with $ac + bc + ad + bd,$ which is equivalent because ring addition is always commutative. Note however that since multiplication is not assumed commutative we must keep each term’s factors in their proper order, e.g. $bc$ not $cb;$ in general when expanding a product $(a_1+a_2+\cdots+a_m) \times (x_1+x_2+\cdots+x_n)$ we get a sum of $mn$ terms $a_i x_j$ whose factor $a_i$ from the first sum must precede the factor $x_j$ from the second sum.
• The facts stated in Example 2, pages 226–227 about zero divisors and units in ${\bf Z} / n {\bf Z},$ which the text says will be proved in the next chapter, are (either explicitly or thinly disguised) facts from the introductory “Chapter 0” that we have already used when finding generators of a cyclic group $Z_n$ and more generally the orders of elements of $Z_n$. Indeed we can say the same for the fact that ${\bf Z} / n {\bf Z}$ is a ring in the first place (Example 4 on p.224; recall also the digression in class about “casting out nines” a few weeks ago).
• We will not say much about the details of quadratic fields ${\bf Q}(\sqrt D)$ and their integer rings, to which the textbook devotes several pages at the end of this first section of Chapter 7. You will learn much more about these intricacies and their ramifications when/if you take our number theory course, Math 124. In the proof that if $D \in {\bf Q}$ is not a square then ${\bf Q}(\sqrt D)$ is indeed a field (p.227, Example 5), you might recognize the key step as the process of “rationalizing the denominator” if that was part of your precollege math curriculum.
» Rings are often denoted by $A$, from French anneau (as in “annulus” etc.) — or $R$ which is probably not from English but the German word, which is also Ring. The German and French words for “field” in the sense of abstract algebra are corps and Körper, each having “body” as its primary meaning, and coincidentally each being its own plural; the German word must be why a field is often called $K$. Curiously the French for “division ring“ = “skew field” is corps gauche which is literally “left field”.
• (covered November 9) [...]
• It is not too hard to show that if there exists a finite field $F$ of $q$ elements then $q$ is a prime power, say $q = p^e$ ($e \geq 1),$ and the additive group of $F$ must be an “elementary abelian $p$-group” i.e. $(F,+) \cong ({\bf Z} / p{\bf Z})^e.$ It is rather harder (except for $e=1)$ to show that conversely for each prime power $q$ there exists a finite field $F$ of $q$ elements, and moreover $F$ is unique up to isomorphism. This will be a key result in Math 123, where you will also learn that ${\rm Aut}(F)$ is a cyclic group of order $e$ generated by the Frobenius map $x \mapsto x^p.$

November 9: Further examples of rings (D&F 7.2)

Remarks on D&F 7.2
• [...]
• If $R$ does have zero divisors then $R[x]$ always has zero divisors (because it contains $R$ as a subring), and may also have nonconstant units — for example, if $R = {\bf Z} / 4 {\bf Z}$ then $(\bar{1}+\bar{2}x) (\bar{1}+\bar{2}x) = \bar{1}+\bar{4}x+\bar{4}x^2 = 1 + 0 + 0 = 1.$
• Unlike a polynomial ring $R[x]$, a group ring $R[G]$ always has units not in $R,$ unless $G$ is the trivial group when $R[G]$ contains nothing but $R$: if $g \in G$ then $g,$ considered as the element $1_R g$ of $R[G],$ is a unit with two-sided inverse $1_R g^{-1}.$ Moreover $R[G]$ can have zero divisors even when $R$ is a field. For example, if $G$ contains an element $x$ of order 2 then $(1_G+x) (1_G-x) = 1-x^2 = 1-1 = 0$ in $R[G].$ Can you generalize this example to any $x \neq 1_G$ of finite order?
• When working with group rings $R[G]$ we always write the group $G$ multiplicatively even when the group is abelian. The group ring $R[G]$ can be defined more generally without the hypothesis that $G$ be finite, as long as we allow only finite sums $\sum_{i=1}^n a_i g_i$ ($a_i \in R, g_i \in G$); for example that is how Wikipedia’s “group ring” article defines it. An important special case is $G = {\bf Z}.$ Since we must write $G$ multiplicatively, we describe it as an infinitely cyclic group $\langle x \rangle.$ Then $R[G]$ consists of “$2$-sided polynomials” $\sum_{i=-m}^n a_i x^i$ and thus contains a copy of the polynomial ring $R[x]$, namely the subring consisting of sums $\sum_{i=-m}^n a_i x^i$ where $a_i = 0$ for all $i<0.$ Generalizing the description of units in $R[x],$ we readily see that if $R$ is a ring with unit and no zero divisors then the only units in $R[{\bf Z}]$ are monomials $a x^i$ with $a \in R^\times$ and $i \in {\bf Z}.$
» Logically a polynomial such as $a x^i$ ($a \neq 0$) with a single term should be called a “mononomial” (cf. binomial, trinomial, and indeed polynomial). But when two consecutive sounds or syllables are identical or nearly so, one may be lost in a process ironically named haplology (the first part is from the same Greek root as haploid).

November 14: MIDTERM REVIEW

November 16: MIDTERM

Tenth problem set: Introduction to rings

November 21: Digression on projective linear groups, projective spaces, and projective geometry

NO CLASS NOVEMBER 23 (Thanksgiving recess)

November 28: Introduction to vector spaces (D&F 11.1; cf. 10.1 and 10.2)

Remarks on D&F 11.1
• Already in Proposition 1 we see that the existence of inverses for every nonzero element of $F$ constrains the structure of $F$-vector spaces compared with modules over an arbitrary commutative ring with 1, or even over an integral domain. For example, $\bf Z$ considered as a module over itself (a.k.a. an infinite cyclic group) is generated by $\{2,3\},$ and neither $2$ nor $3$ can be removed while still generating all of $\bf Z$, but they are not “$\bf Z$-linearly independent” (for example, $6 \cdot 2 + (-4) \cdot 3 = 0$).
• One can also define vector spaces over $F$ when $F$ is a (not necessarily commutative) division ring; the basic results, at least through 11.2, hold in this context too (with some care to distinguish left vector spaces from right vector spaces; in 11.2, a left vector space consists of column vectors, and a right vector space consists of row vectors). We will not pursue this further here, for two reasons. First, though vector spaces over division rings are a natural generalization, they arise much more rarely in practice than vector spaces over fields (which are ubiquitous in mathematics and the mathematical sciences). Second, the further development of linear algebra, such as connections between eigenvalues and the characteristic polynomial, no longer work when $F$ is not commutative, for lack of a determinant function on square matrices with entries in $F$ (a.k.a. linear transformations from a finite-dimensional $F$-vector space to itself).
• The text’s statement of the “Replacement Theorem” (Thm. 3 on page 410) is a bit confusing; better to say: if ${\cal A}$ is a basis for $V$ with $|{\cal A}| = n$ then there is an ordering $(a_1,a_2,\ldots,a_n)$ of $\cal A$ such that for each $k \in \{1,2,\ldots,m\}$ the $n$-tuple $(b_1,b_2,\ldots,b_k,a_{k+1},a_{k+2},\ldots,a_n)$ is an (ordered) basis of $V$. [There is also a tiny typo at this point in D&F: the final “V” should be $V,$ in italic font, since it's a mathematical formula; nowadays this might be called a “TeX-o”, but D&F predates (La)TeX.]

November 30: Vector spaces cont’d (D&F 11.1); linear transformations and their matrices (D&F 11.2, omitting tensor and Kronecker products)

Further remarks on D&F 11.1
• The text’s example of the vector space of solutions of $y'' - 3 y' + 2 y = 0$ requires some further explanation: yes, it comes down to the linearity of differentiation, but the details are not immediate. To put $y'' - 3 y' + 2y = 0$ in the context of linear algebra, the three functions $y,y',y''$ must be in the same vector space of functions, call it $V;$ and differentiation should be a linear transformation from $V$ to itself so that it makes sense to iterate it (getting from $y$ to $y'$ to $y''$) and to form the linear combination $y'' - 3y' + 2y$ of $y,y',y''.$ One way to do this is to let $V$ be the ${\bf R}$-vector space of infinitely differentiable functions from $\bf R$ to $\bf R$. Then differentiation is a map $D: V \to V,$ and the linearity of $D$ amounts to familiar rules of differential calculus: $(y_1 + y_2)' = y_1' + y_2'$ for all $y_1,y_2\in V,$ and $(ay)' = a y'$ for all $a \in \bf R$ and $y \in V.$ Since the composition of linear transformations is again linear we automatically get linearity of the map $D^2 = DD$ taking any $y \in V$ to its second derivative $y''$. Then the fact that ${\rm Hom}(V,V)$ (or even ${\rm Hom}(V,W)$) itself has the structure of a vector space lets us write $y'' - 3 y' + 2 y = DD y - 3 D y + 2 y$ as the image of $y$ under some linear transformation $T : V \to V,$ namely $T = D^2 - 3D + 2I$ where $I$ is the identity map. Finally $y$ is a solution of $y'' - 3 y' + 2 y = 0$ iff $Ty = 0,$ that is, iff $y$ is in the kernel of $T$ — which we already know is a vector subspace of $V$.
• [What about the image and kernel of $D$ itself? The image is $V$ itself: the preimages of any $y \in V$ are the “indefinite integrals” of $y$. The kernel consists of the constant maps; that accounts for the “constant of integration” in the indefinite integral. In particular $D$ is surjective but not injective. We already know that there are group homomorphisms $G \to G$ that are surjective but not injective or vice versa, necessarily with $G$ infinite; examples such as $D$ show that the analogous phenomenon can happen for linear transformations from a vector space $V$ to itself, necessasrily with $V$ infinite dimensional.]
• Theorem 7 (page 412), the identity $\dim V = \dim W + \dim V/W$ (when $W$ is a subspace of $V$), is the vector-space analogue of the identity $|G| = |H| \cdot |G/H|$ for a group $G$ with a normal subgroup $H$ (and more generally of $|G| = |H| \cdot |G:H|$ when the subgroup is not necessarily normal). Indeed if $F$ and $\dim V$ are finite, with $|F|=q,$ then Theorem 7 is a special case, because then $|V| = q^{\dim V}$ and likewise for $|W|$ and $|V/W|$ (note that $V/W$ is also the quotient of the abelian group $V$ by its subgroup $W$ which is automatically normal).
• Likewise parts 1–3 of Corollary 9 (page 412) are the vector-space analogue of the fact that if $X$ and $Y$ are finite sets of the same cardinality then a map $X \to Y$ is bijective iff it is injective iff it is surjective.
• Even if you haven’t learned of quotient spaces before, you have probably seen Corollary 8 (page 413); it is often called the “rank-nullity theorem” (see the Definition in the middle of this page for the terms “rank” and “nullity”).
• The formula $(q^n-1) (q^n-q) \cdots (q^n-q^{n-1})$ for the number of ordered bases of $F^n$ and the size of ${\rm GL}_n(F)$ (where $F$ is a finite field of size $q$) is analogous to the number $n \cdot (n-1) \cdots 1 = n!$ of permutations of $n$ objects and the size of the symmetric group $S_n$. Likewise the formula $$ \frac{ (q^n-1) (q^n-q) \cdots (q^n-q^{k-1}) } { (q^k-1) (q^k-q) \cdots (q^k-q^{k-1}) } $$ for the number of $k$-dimensional subspaces of $F^n$ is analogous to the formula $$ \frac{ n(n-1)\cdots(n-k+1) }{ k (k-1) \cdots 1 } = \frac{n!/(n-k)!}{k!} $$ for the number $n \choose k$ of $k$-element sets of an $n$-element set. Like the formula for ${n \choose k},$ the count of $k$-dimensional subspaces of $F^n$ is symmetrical under $k \leftrightarrow n-k;$ this isn’t too hard to check from the formula, but a direct explanation is a bit subtler because it's not simply a matter of taking the complement. We shall give a better explanation in Section 11.3 which we cover in the next (and final) lecture.

Remarks on D&F 11.2
• The change-of-basis formula $P^{-1} \! A P$, together with multiplicativity of the determinant, implies that $\det A$ is coordinate-independent and thus well-defined on the endomorphism ring ${\rm End}(V) = {\rm Hom}(V,V)$. More generally, so is the characteristic polynomial $\det(\lambda I - A).$

December 5: Duality (D&F 11.3, omitting most of the infinite-dimensional case)

Remarks on D&F 11.3
• You might wonder why we don’t devote such attention to ${\rm Hom}(F,V),$ which likewise has the same dimension as $V$ when $V$ is finite dimensional. The reason is that this space is easily identified with $V,$ for any vector space $V,$ and not even needing to construct a basis; indeed it works even if we require only that $F$ be a ring with $1$ and that $V$ be an $F$-module. Namely, there is an isomorphism $i: {\rm Hom}(F,V) \to V,$ defined by $i(w)=w(1).$ The inverse isomorphism takes any $v \in V$ to the linear transformation $F \to V$ taking any $c \in F$ to $cv.$
• When $V$ and $W$ are finite-dimensional, this is also consistent with our identification of ${\rm Hom}(V,W)$ with a vector space of matrices once we choose bases for $V$ and $W$. In this picture, elements of $V$ and $W$ are identified with column vectors. If $V = F$ then ${\rm Hom}(V,W)$ becomes the space of matrices with a single column of length $\dim W$ — so we might not be too surprised that it is basically the same as $W$. But if $W=F$ then ${\rm Hom}(V,W)$ becomes the space of matrices with a single row of length $\dim V$ — and that turns out to be quite different. Indeed if we apply some invertible matrix $A$ to change the basis of $V$ then the associated dual basis of $V^*$ changes not by $A$ but by the inverse transpose $(A^{-1})^{\sf T} = (A^{\sf T})^{-1}.$ If $u \in V^*$ is represented by some row vector $(u_1,\ldots,u_n)$ then, for any column vector $v = (v_1,\ldots v_n)^{\sf T}$ representing a vector in $V$, the scalar $u(v)$ is the “dot product” $u_1 v_1 + \cdots + u_n v_n;$ this again is consistent with matrix multiplication: the product of a one-row and one-column matrix of the same size is the $1 \times 1$ matrix consisting of the dot product of that row and that column. Also, $V^{**}$ then corresponds to transposed row vectors, which are again column vectors, allowing a natural identification of $V$ with its second dual.
• Suppose $f : U \to W$ is the composite of linear transformations $U \stackrel{\psi}{\to} V \stackrel{\varphi}{\to} W$. What then is $f^* : W^* \to U^*?$ As you might expect from the matrix sizes (in the finite-dimensional case), $f^*$ is the composit of $\psi^*$ with $\varphi^*$ in reverse order. This corresponds to the matrix identity $(AB)^{\sf T} = B^{\sf T} A^{\sf T}.$ A common way to visualize this in the diagram is to say that duality reverses the arrows: $U^* \stackrel{\;\psi^*}{\leftarrow} V^* \stackrel{\;\varphi^*}{\leftarrow} W^*.$
• So far in the class we never mentioned or used the Axiom of Choice (or equivalently Zorn’s Lemma), though that too is in D&F’s “Chapter Zero”. In particular, we did not claim that even infinite-dimensional vector spaces $V/F$ always have bases, and thus did not prove that every linearly independent set in $V$ can be completed to a basis. (Even the example $(F,V)=({\bf Q},{\bf R})$ would require an uncountable basis $B$ that would be impossible to exhibit in any tangible way, even though the existence of such a “Hamel basis” $B$ can be proved assuming the Axiom of Choice.) Thus we will not be able to prove most of the statements in 11.3 about infinite dimensional vector spaces. However, we can still obtain some results about vector spaces with a countable basis, such as $F[x]$ — see the final exercise.

Eleventh and last problem set: Introduction to modules, vector spaces, and linear transformations
(5i) corrected thanks to Alex Rojas and, indepedently, Jeffery Wang

December 7: REVIEW

Wednesday, December 13:
FINAL EXAMINATION
2–5 PM in Science Center D