MATH 126 SOLUTION SET 2

NILS R. BARTH

1. To avoid confusion, I’ve changed the indices to range from 1 to n, rather than 1 to m^{1}. First part:

• R_{i}M = e_{i}M: e_{i} ∈ R_{i}, so e_{i}M ⊂ R_{i}M; r_{i}m = e_{i}r_{i}m, so R_{i}M ⊂ e_{i}M

• M = ∑ e_{i}M: m = 1m = (e_{1} + e_{2} + ⋯ + e_{n} )m = e_{1}m + e_{2}m + ⋯ + e_{n} m

• M = ⊕e_{i}M: e_{i}m = m for m ∈ e_{i}M, and e_{i}m = e_{i}e_{j}m = 0m = 0 for m ∈ e_{j}M, i ≠ j Second part: Recall that R_{i} = ∑ L_{i}, for any left ideal L_{i}. Any module over a semisimple ring is

semisimple, so M = ⊕M_{i}, where M_{i} is simple. Recall that

L_{i}M_{j} = / \begin{matrix}0 & M_{j} ≇ L_{i} \\ L_{i} & M_{j} ≅ L_{i}. \end{matrix} (2.1)

Thus,

R_{i}M = ∑ L_{i} ⊕ M_{j} = ∑ L_{i}M_{j} = ∑_{N≅L_{i}} N (2.2)

Note: this is Theorem XVII.4.4 in Lang’s Algebra.

2. Let us present D_{4} = ⟨σ, τ | σ^{4} = τ^{2} = 1, στ = τσ^{−1}⟩ . By the previous solution set, the character table for D_{4} is:

1 σ^{2} σ, σ^{−1} τσ^{2} τσ, τσ^{−1}
χ_{1} 1 1  1  1  1
χ_{2} 1 1  1 −1  −1
χ_{3} 1 1 −1  1  −1
χ_{4} 1 1 −1 −1  1
χ_{5} 2 −2  0  0  0

Character Table 1: D_{4}

All elements of O either commute or anti-commute. Thus, the conjugacy classes of O are {1}, {−1}, {±i} , {±j}, {±k}. Clearly, ⟨i⟩ , ⟨j⟩, ⟨k⟩ are all normal subgroups, so we can map any of them to zero and we are left with C_{2}, whose representations come from sending its generator to ±1. Quotienting by i leaves generator j = k, so j, k are sent to the same element. Similarly for quotienting by j, k. Omitting the repeated trivial representation, we obtain the following partial character table for O:

1 −1 ±i ±j ±k
χ_{1} 1 1 1 1 1
χ_{2} 1 1 1 −1 −1
χ_{3} 1 1 −1 1 −1
χ_{4} 1 1 −1 −1 1
χ_{5}

Partial Character Table 1: O

Received by the editors October 14, 1998.

1Since m will be used to denote elements of M.

The last row is uniquely determined by its dimension (which must be 2), and orthogonality^{2}. Since the first four rows agree with the character table for D_{4} and the conjugacy classes have the same size, the last row must agree as well. Thus, O has character table:

1 −1 ±i ±j ±k
χ_{1} 1 1 1 1 1
χ_{2} 1 1 1 −1 −1
χ_{3} 1 1 −1 1 −1
χ_{4} 1 1 −1 −1 1
χ_{5} 2 −2 0 0 0

Character Table 2: O

Last, we must verify that D_{4} ≇ O. This is because the former has two elements of order 4, while the latter has six. Thus, non-isomorphic groups can have the same character table.

3. The character table for S_{4} is:

1 (2) (3) (4) (2, 2)
χ_{1} 1 1 1 1  1
χ_{2} 1 −1 1 −1  1
χ_{3} 2 0 −1 0  2
χ_{4} 3 1 0 −1 −1
χ_{5} 3 −1 0 1 −1

Character Table 3: S_{4}

The first two representations are the trivial and sign, the fourth comes from decomposing the permutation representation: χ_{π} = χ_{1} ⊕ χ_{4}, the fifth from tensoring the sign and fourth: χ_{5} = χ_{1} ⊗ χ_{4}, and the third follows from dimension and orthogonality.

4. The commutator of S_{n} is A_{n}, so the one dimensional representations of S_{n} correspond to representations

of S_{n} /A_{n} = C_{2}, and are thus the trivial and sign representations.

S_{1}, S_{2} are commutative, so they have no simple two dimensional representations, S_{3} ≅ D_{3}, and its simple two dimensional representation is on solution set 1.

For S_{4}, note that the double conjugations (the elements of the form (ab)(cd)) form a normal subgroup of order 4 (in fact the Klein 4-group), so quotienting out by them gives a group of order 24/4 = 6, which is non-commutative, as we haven’t quotiented out by A_{n}. Thus, the quotient is S_{3}, and composing the quotient with the simple representation of S_{3} yields a simple representation of S_{4}. By inspecting the character table, this is the only simple two dimensional representation.

For n > 4, S_{n} has no proper normal subgroups other than A_{n}. Thus, the kernel of any representation must be 1, A_{n} , S_{n}. The latter two correspond to the sign and trivial representations. Thus, any other simple representation must be faithful.

Take two disjoint 2-cycles, say (12), (34). These are commuting and diagonalizable, and are thus simultaneously diagonalizable. Thus, ρ(12) = ( \begin{matrix}±1 & 0 \\ 0 & ±1 \end{matrix} ) and similarly for (34). However, they cannot be sent to ±I, as they then would commute with everything and this representation would not be faithful. In particular, if (12) is sent to ±I, then ρ(1234) = ρ((12)(1234)(12)) = ρ(2134) and the representation will not be faithful. The same holds for (34) and (12)(34). Thus, ρ((12)) = ± ( \begin{matrix}1 & 0 \\ 0 & −1 \end{matrix} ) and similarly for (34). But then ρ((12)(34)) = ±I (since they are simultaneously diagonalizable), which cannot happen, by the above reasoning. Thus, for n ≥ 5, S_{n} has no simple two dimensional representations.

The above can be generalized. What we did was find a commutative subgroup of S_{n}, all of whose elements have order 2. Thus, their image is simultaneously diagonalizable, and there are thus 2^{d} possible elements of the image (d is the dimension of the representation). However, none of these elements can be −I, as this representation would then fail to be faithful. Since we can always find r = ⌊\frac{n}{2} ⌋ disjoint 2-cycles, the subgroup they generate has order 2^{r} elements, which must be mapped injectively into a set of size 2^{d} −1. Thus, for n ≥ 5, simple representations of S_{n} (other than the trivial and sign) must have dimension > ⌊\frac{n}{2} ⌋.

2Actually, it is uniquely determined by orthonormality and the need for the first entry to be positive.