TRANSLATING THE IRREDUCIBLE REPRESENTATIONS OF S_{n} INTO GL_{n}(F_{q})

DAVID G. HARRIS

1. Introduction

1.1. The Argument. In both S_{n} and GL_{n}(F_{q}), we send letters (respectively, basis elements) to other letters (basis elements). In GL_{n} we can take linear combinations, which we can’t do in S_{n}; but this is really the only difference. GL_{n}, loosely speaking, is an obese, redundant version of the svelte S_{n}.

The goal of this paper is to make this correspondence precise, to show that, under this correspondence between letters and bases, we can import many of GL_{n}’s representations wholesale from S_{n}.

Section 2 is based on [2]. Section 3 is based on [1].

1.2. Permutation Representations in S_{n}. To begin, we clarify the representations of S_{n}. All S_{n}’s representations begin with the permutation representations; though these aren’t themselves irreducible, we can construct all the irreducibles from them. We saw in class the simplest kind of permutation representation, the representation of how a permutation acts on single letters. We can look at a more general kind of representation, one acting on a group of letters simultaneously.

More precisely, for each partition μ of n, we define tableaux: write a table with μ_{1} columns in the first row, μ_{2} columns in the second row, and so on. In each cell, put one number 1, 2, …, n, each appearing exactly once in the diagram.

Here is a tableau of type (1, 2): _{2 3}^{1}

Here is different tableau of type (1, 2): _{3 2}^{1}

Here is a tableau of type (1, 4, 5): 5 2 1 9 10 3 4 6 7 8

This isn’t quite the concept we want to use for defining the permutation representations of S_{n}. Instead, let’s define a tabloid as a tableau modded out by rearranging the letters in a row. So, as tabloids, _{2 3}^{1} is the same as _{3 2}^{1} , but different from _{1 3}^{2} .

We put a bar over tabloids to show that you can rearrange the letters in any row. As notation, we put {t} to be t’s associated tabloid.

Now, for each partition μ, S_{n} has a permutation representation on tabloids of type μ, with the natural action.

2. Some irreducibles of GL_{n}

The corresponding concept in GL_{n} is flags. Given a partition μ = 0 ≤ μ_{1} ≤ μ_{2} ≤ μ_{n} of n, we define a flag e(μ) of type μ as a collection of subspaces V_{1} ⊂ V_{2} ⊂ ⋯ ⊂ V_{m}, where dim V_{i} = μ_{1} +⋯+μ_{i}. That is, e(μ) is some flag of the same shape as our partition.

Since all g ∈ GL_{n} are bijections, dim g(V_{i}) = dim V_{i}; i.e., g permutes flags of type μ. So we can consider the permutation character ρ(μ) for flags of type μ.

Received by the editors February 28, 1999.

Corresponding to this is R(μ), the permutation character of S_{n} on tabloids of type μ.

Before we compute our irreducibles, we need to prove two linear-algebra lemmata.

Lemma 2.1. Let e and f be two flags of type \overbrace{(1, 1, …, 1)}^{n} = 1^{n}. Then we can choose vectors v_{1}, …, v_{n}, such that ⟨v_{1}, …, v_{i}⟩ = e_{i} and ⟨v_{π(1)}, …, v_{π(i)}⟩ = f_{i} for some permutation π.

Proof. By induction on n. If n = 1, this is trivial: e = f.

If n > 1, proceed as follows: Note that e′ = e ∩ e_{n−1} and f′ = f ∩ e_{n−1} are both flags of type 1^{n−1}. (By f ∩ e_{n−1}, we mean the flag f_{1} ∩ e_{n−1} ⊂ f_{2} ∩ e_{n−1}. This is a flag of type 1^{n−1} since f ends at f_{n} ∩e_{n−1} = V ∩e_{n−1} = e_{n−1}, which has dimension n − 1, and at each step f_{i} ∩ e_{n−1} can only increase in dimension by one, since f_{i} does so.)

So, by induction, we can choose vectors v_{1}, … v_{n−1} so that ⟨v_{1}, …, v_{i}⟩ = e′_{i} and ⟨v_{π(1)}, …, v_{π(i)}⟩ = f′_{i}.

Choose z so that f_{z} ⊂ e_{n−1} but f_{z+1} ⊄ e_{n−1}. (There must be some such z, since f(0) = ∅ ⊂ e_{n−1} but f(n) = V ⊄ e_{n−1}.) Now, choose v_{n} to be any element of f_{z+1} \ f_{z}.

First, ⟨v_{1}, …, v_{i}⟩ = e_{i}:

For i < n, there is no problem: ⟨v_{1}, …, v_{i}⟩ = e′_{i} = e_{i}.

Note, since v_{n} ∉ f_{i}, e′_{z+1} = f_{z} ⊕ v_{n}. Since f_{z} ⊂ e_{n−1} but f_{z+1} ⊄ e_{n−1}, it must be that v_{n} ∉ e_{n−1}. So ⟨v_{1}, …, v_{n}⟩ = e_{n−1} ⊕ v_{n} = V, as desired.

Second, with the order

v_{1}, v_{2}, …, v_{π(z)}, v_{n}, v_{π(z+1)}, … v_{π(n−1)},

the v_{i} span f too:

For i ≤ z, ⟨v_{π(1)}, …, v_{π(i)}⟩ = f′_{i} = f_{i} ∩ e_{n−1} = f_{i} since f_{i} ⊂ e_{n−1}.

For i = z + 1: ⟨v_{π(1)}, …, v_{π(z)}, v_{n}⟩ = f_{z} ⊕ v_{n} = f_{z+1}, since v_{n} ∈ f_{z+1} \ f_{z}.

For i > z + 1: ⟨v_{π(1)}, …, v_{π(z)}, v_{n}, v_{π(z+1)}, … v_{π(i)}⟩ = f_{i−1} ⊕ v_{π(i)}, By induction, v_{π(i)} ∈ (f_{i} ∩ e_{n−1}) \ (f_{i−1} ∩ e_{n−1}). So, v_{π(i)} ∈ f_{i} \ f_{i−1}, as desired.

As a corollary, given any two flags e(μ) and f(λ), we can choose vectors v_{1}, …, v_{n} and a permutation π such that, for all i, ⟨v_{1}, …, v_{μ_{1}+···+μ_{i}}⟩ = e_{i} and ⟨v_{π(1)}, …, v_{π(λ_{1}+···+λ_{i})}⟩ = f_{i}. Just expand e and f to flags of type 1^{n}: for example, expand F_{q} ⊂ F_{q}^{3} to F_{q} ⊂ F_{q}^{2} ⊂ F_{q}^{3}. Then choose vectors and a permutation as above.

Lemma 2.2. For any partitions μ, λ, the number of orbits of pairs (e(μ), f(λ)) under GL_{n} is equal to the number of the orbits of pairs (s(μ), s(λ)) in S_{n}. By s(μ), we mean a tabloid of type μ.

Proof. By the previous lemma, we can choose vectors v_{1}, … v_{n} and a permutation π such that the ⟨v_{1}, …, v_{μ_{1}}⟩ = e(μ)_{1}, and so forth.

To each (e(μ), f(λ)), associate the orbit of

( 1  ... μ_{1} ; μ_{1}+1  ... μ_{1}+μ_{2} ; ⋮ ; μ_{1}+···+μ_{n−1}+1 ... n , π(1)  ··· π(λ_{1}) ; π(λ_{1}+1)  ··· π(λ_{1}+λ_{2}) ; ⋮ ; π(λ_{1}+···+λ_{n−1}+1) ··· π(n) ) .

This association is injective on orbits—if (e(μ), f(λ)) and (e′(μ), f′(λ)) both go to the same pair, then define g to be the change-of-basis matrix between the two different choices of vectors v: this will carry (e, f) to (e′, f′), so they are in the same orbit.

It is well-defined on orbits—if g carries (e, f) to (e′, f′), then if we choose v_{1}, … v_{n} for e and f, then gv_{1}, …, gv_{n} span e′ and f′ in the desired way, so (e′, f′) is associated with the same orbit of S_{n}.

It is surjective—we can achieve any orbit associated with π, λ, and μ by choosing e = ⟨ϵ_{1}, …, ϵ_{μ_{1}}⟩ ⊂ ⟨ϵ_{1}, …, ϵ_{μ_{1}+μ_{2}}⟩ ⊂ ⋯, and f = ⟨ϵ_{π(1)}, …, ϵ_{π(λ_{1})}⟩ ⊂ ⟨ϵ_{π(1)}, …, ϵ_{π(λ_{1}+λ_{2})}⟩ ⊂ · · ·.

Now that we have proved the second lemma, we can define what our irreducible character χ_{μ} of GL_{n} shall be.

Given a partition μ, put μ′ = (μ_{1} + 1, μ_{2} + 2, μ_{3} + 3, …, μ_{n} + n). Then define χ_{μ} = ∑_{π∈S_{n}} sgn(π)ρ(μ′ − π).

By μ′ − π we mean the term-by-term subtraction (μ′_{1} − π(1), μ′_{2} − π(2), …, μ′_{n} − π(n)).

Theorem 2.3. The characters χ_{μ} are distinct irreducible characters of GL_{n}(F_{q}).

Proof. First, note that the χ_{μ} are integral linear combinations of characters. So, we need to show (χ_{μ}|χ_{λ}) = δ(μ, λ), and that χ_{μ}(1) > 0.

Note that if r is a transitive, faithful representation, then (r|1) is the number of its orbits: for each orbit {x_{1}, …, x_{k}}, x_{1} + ⋯ + x_{k} is fixed by r, hence provides a stable one-dimensional subspace, and conversely so.

Using Lemma 2.2:

(ρ(μ)|ρ(λ)) = (ρ(μ) ⊗ ρ(λ)|1)

= # orbits of flags (e(μ), e(λ))

= # orbits of tabloids (s(μ), s(λ))

= (R_{μ} ⊗ R_{λ}|1)

= (R_{μ}|R_{λ}).

We can disregard the complex conjugation since permutation representations are defined over R.

We can apply this formula to calculate (χ_{μ}, χ_{λ}):

(χ_{μ}|χ_{λ}) = \frac{1}{|GL_{n}|} ∑_{x∈GL_{n}} ∑_{π} sgn(π)ρ(μ′ − π, x) ∑_{σ} sgn(σ)ρ(λ′ − σ, x)

= ∑_{π,σ} sgn(π) sgn(σ) [\frac{1}{|GL_{n}|} ∑_{x∈GL_{n}} ρ(μ′ − π, x)ρ(λ′ − σ, x)]

= ∑_{π,σ} sgn(π) sgn(σ)(ρ(μ′ − π)|ρ(λ′ − σ))

= ∑_{π,σ} sgn(π) sgn(σ)(R(μ′ − π)|R(λ′ − σ))

= (∑_{π} sgn(π)R(μ′ − π)| ∑_{σ} sgn(σ)R(λ′ − σ))

= (ϕ(μ)|ϕ(λ))

= δ(μ, λ).

Here ϕ(μ) is the character ∑_{π} sgn(π)R(μ′ − π) of S_{n}. The last line holds because the ϕ are irreducible in S_{n}. We will not prove this formula in this paper.

So, the χ’s are irreducible and distinct.

This computation is very powerful. It not only shows that the χ’s are irreducible, but that they act just like the R’s. They have the same formula in terms of permutation representations; χ’s dimensions are given by similar formulas (see below). In fact, we might call χ a quantized version of R. In the limit as q goes to 1, the χ become R.

Let’s look at an example.

Proposition 2.4. This formula holds for the partition μ = (0, 1, 2) of GL_{n}(F_{q}^{3}).

Proof. We have

ϕ_{μ} = ∑_{π∈S_{3}} ρ(1, 3, 5 − π)
= ρ(0, 1, 2) − ρ(0, 0, 3)

The permutation representation ρ(0, 1, 2) is just the permutation representation on the standard flag spanned by ϵ_{1}, ϵ_{2}, ϵ_{3}, and ρ(0, 0, 3) is trivial (there is only one flag of type (0, 0, 3)). If we check the character with the inner product, we see that this representation is, indeed, simple.

We have postponed the proof that χ(μ, 1) > 0, as it is technical. This follows in a separate theorem.

Theorem 2.5. The χ_{μ} are bonafide characters, not just virtual characters.

Proof. Now, we need to show χ(μ, 1) > 0. This depends on two technical calculations:

First, let’s compute the dimension of ρ(μ): this is the number of flags of type μ. Note for every orthonormal choice of basis vectors ϵ_{1}, … ϵ_{n}, we can associate the

corresponding flag ⟨ϵ_{1}, …, ϵ_{μ_{1}}⟩ ⊂ ⟨ϵ_{1}, …, ϵ_{μ_{1}+μ_{2}}⟩ ⊂ ⋯ ⊂ ⟨ϵ_{1}, …, ϵ_{n}⟩.

If we denote by [i]! the number of orthonormal ways of choosing a basis for an i-dimensional vector space, we have [n]! flags so far.

However, this correspondence isn’t unique. Any such choice of basis vectors e_{1}, …, e_{μ_{1}} gives us the same V_{1}. Similarly, given our choice of e_{1}, …, e_{μ_{1}}, we can choose orthonormal vectors for V_{2}/V_{1}, i.e., [μ_{2}]! more choices.

Continuing this way, we see the dimension of ρ(μ) is

\frac{[n]!}{[μ_{1}]! ⋯ [μ_{n}]!}.

Now, let’s calculate [i]!. We can choose an orthonormal basis for V of dimension i by first choosing any unit vector v in V, then choosing a vector in v^{⊥}, and so on. So [i]! = [i][i − 1] … [1], where [i] is the number of unit vectors in an i dimensional vector space of F_{q}.

In F_{q}^{i}, there are q^{i}−1 non-zero vectors, each of which has q−1 distinct multiples, for (q^{i} − 1)/(q − 1) unit vectors. So [i] = q^{i−1} + q^{i−2} + ⋯ + 1.

(A bit of notation: We define [i]! = 0 for i < 0.)

Note the correspondence of this formula with the formula for the dimension of R(μ), ρ’s S_{n}-counterpart.

Now, we can calculate χ_{μ}(1):

χ_{μ}(1) = ∑_{π} sgn(π)ρ(μ′ − π, 1)

= ∑_{π} sgn(π)\frac{[n]!}{([μ′_{1} − π_{1}]!) ⋯ ([μ′_{n} − π_{n}]!)}

= [n]! det |([μ′_{i} − j]!)^{−1}|

= [n]! det |\frac{1}{[μ′_{i} − j][μ′_{i} − (j + 1)] ⋯ [1]}|

= [n]! det |\frac{[μ′_{i}] ⋯ [μ′_{i} − (j − 1)]}{[μ′_{i}]!}|

= \frac{[n]!}{[μ′_{1}]! ⋯ [μ′_{n}]!} det |[μ′_{i}][μ′_{i} − 1] ⋯ [μ′_{i} − (j − 1)]|

= \frac{[n]!}{[μ′_{1}]! ⋯ [μ′_{n}]!} det |[μ′_{i}]\frac{[μ′_{i}] − [1]}{q} ⋯ \frac{[μ′_{i}] − [j − 1]}{q^{j−1}} |

The last line follows from the identity [a − b] = ([a] − [b])/q^{b}.

To calculate this determinant, we’ll substitute indeterminates X_{i} for [μ′_{i}]:

det |X_{i} \frac{X_{i} − [1]}{q} … \frac{X_{i} − [j − 1]}{q^{j−1}} | .

The degree of this is n(n − 1)/2 as a polynomial in the X_{i}: row j’s cells each have degree j−1, so when we calculate the determinant, each summand will have degrees 0 + 1 + 2 + ⋯ + (n − 1). Also, if X_{i} = X_{k}, then columns i and k will be the same, and the determinant will vanish. So, this determinant is a constant multiple of the van der Monde determinant ∏_{i<j}(X_{j} − X_{i}).

Let’s find what this constant multiple is. We need to find the coefficient of one term. The coefficient of the X_{2}X_{3}^{2}X_{4}^{3} ⋯ X_{n}^{n−1} term is this: in taking the determinant, we must choose the cells (1, 0), (2, 1), and so on, since column determines which X_{i} and row determines its degree. This gives coefficients

1 × 1/q × 1/q^{1+2} × 1/q^{1+2+3} × ⋯

which equals q^{−n(n−1)(n−2)/3}, whereas in the van der Monde determinant, this term would have coefficient unity. So

det |X_{i} \frac{X_{i} − [1]}{q} ⋯ \frac{X_{i} − [j − 1]}{q^{j−2}} | = q^{N} ∏_{i<j}(X_{j} − X_{i}).

Whew! Almost done.

Let’s plug this into our formula for χ(μ, 1):

χ(μ, 1) = \frac{[n]!q^{N} ∏_{i<j}([μ′_{j}] − [μ′_{i}])}{[λ_{1}]! ⋯ [λ_{n}]!}

This is positive. The [x]! terms are all positive, the q^{N} term is positive, and, since the μ′_{i} are increasing (since the μ_{i} are), each ([μ′_{j}] − [μ′_{i}]) term is positive too.

It’s beyond the scope of this paper to prove that

ϕ(μ) = ∑_{π} sgn(π)R(μ + (1, 2, …, n) − π)

is an irreducible character of S_{n}.

We can point the way to this result, by finding a formula for S_{n}’s irreducibles.

3. The irreducible representations of S_{n}

The natural place to start is with the permutation representation M_{μ} for a partition μ. (The notation M_{μ} means we consider it as a C[S_{n}]-module, with the natural action of S_{n} on tabloids.) As a module, M_{μ} has basis {t}, where t are tabloids of type μ. We have seen that, convenient though it be, M_{μ} is not simple. For example, if μ is the trivial partition (1, 1) in S_{2}, then {1} {2} + {2} {1} gives a one-dimensional fixed submodule. However, we can find a good submodule of M_{μ} which is irreducible.

Let us define e_{t}, for any tableau t of type μ, to be

e_{t} = {t} ∑_{π∈C_{t}}(sgn π)π.

Here, C_{t} is the column-stabilizer of t, the set of permutations which don’t move letters from one column of t to another. So, for the tableau _{2 3}^{1} , the column stabilizer is {(1 2), 1)}. The addition of tabloids here is purely formal.

e_{t} is called a polytabloid. It is crucial in this definition that, even though e_{t} is a tabloid, t itself is a tableau—that is, the order of the letters in t’s rows is important, but not in e_{t}’s.

Here is an example. If t = 1 2 3 , then e_{t} = 1 2 3 − 2 1 3 . That is, given a tableau t, we permute the columns in all possible ways, and take the sum of these tabloids taking the sign of our permutations into account, and then draw a line on top at the end.

Now, define M′_{μ} to be the submodule spanned by the polytabloids. This is called the Specht submodule.

First, we prove some lemmata.

Lemma 3.1. M′_{μ} is generated by any polytabloid.

Proof. This follows from the fact that e_{t}π = e_{tπ}, which in turn follows from the fact that C_{tπ} = π^{−1}C_{t}π.

Lemma 3.2. Let t and t′ be tableaux of type μ such that {t′} ∑_{π∈C_{t}}(sgn π)π ≠ 0. Then {t′} ∑_{π∈C_{t}}(sgn π)π = ±e_{t}.

Proof. Let a and b be any two numbers in the same row of t′. Say a and b were in the same column of t. Then {1, (a b)} would be a subgroup of C_{t}. Choosing g_{1}, …, g_{l} to be its coset representatives, we would have

{t′} ∑_{π∈C_{t}}(sgn π)π = {t′} ((e − (a b))(g_{1} + ⋯ + g_{k}))

= ({t′} − {t′} (a b))(g_{1} + ⋯ + g_{k})

= ({t′} − {t′})(g_{1} + ⋯ + g_{k})

(Since a, b are in the same row, {t′} (a, b) = {t′}—it’s a tabloid.) = 0.

So, for every i, the numbers in the ith row of t′ lie in in different columns of t. Since {t′} is a tabloid, this means we can rearrange {t′} such that for every letter k, k lies in the same column of {t′} as in t. In other words, {t′} = {t} ρ for some ρ ∈ C_{t}. So

{t′} ∑_{π∈C_{t}}(sgn π)π = {t} ρ ∑_{π∈C_{t}}(sgn π)π

= {t} ∑_{π∈C_{t}}(sgn π)πρ

= ± {t} ∑_{π∈C_{t}}(sgn πρ)πρ

= ± {t} ∑_{πρ∈C_{t}}(sgn πρ)πρ, since ρ ∈ C_{t}

= ±e_{t}.

So, for all tableaux t′, t′ ∑_{π∈C_{t}}(sgn π)π is a multiple (0, 1, or −1) of e_{t}.

Lemma 3.3. If u ∈ M_{μ} and t is a tableau of type μ, then u ∑_{π∈C_{t}}(sgn π)π is a multiple of e_{t}.

Proof. By definition of M_{μ}, u is a linear combination of tabloids {t′}. By the previous lemma, t′ ∑_{π∈C_{t}}(sgn π)π are each multiples of e_{t}.

Now, define a bilinear form ⟨ | ⟩ on M_{μ} by:

⟨{t_{1}} | {t_{2}}⟩ = δ({t_{1}} , {t_{2}}).

This form is visibly symmetric and S_{n}-invariant.

Lemma 3.4. If U ≤ M_{μ}, then either M′_{μ} ⊂ U or U ⊂ M′_{μ}^{⊥}.

Proof. Let u ∈ U. Then, by the previous lemma, for all tableaux t, u ∑_{π∈C_{t}}(sgn π)π = k_{u,t}e_{t}, for k_{u,t} ∈ C. There are two cases:

• For some tableau t and u ∈ U, k_{u,t} ≠ 0. Then we have

e_{t} = \frac{u ∑_{π∈C_{t}}(sgn π)π}{k_{u,t}} ,

so e_{t} ∈ U (U is a submodule). Since M′_{μ} is generated by e_{t}, it must be that M′_{μ} ⊂ U.

• For all tableaux t and u ∈ U, k_{u,t} = 0. Then for all u, t

0 = u ∑_{π∈C_{t}}(sgn π)π

= ⟨u ∑_{π∈C_{t}}(sgn π)π| {t}⟩

= ∑_{π∈C_{t}}⟨uπ|(sgn π) {t}⟩

= ∑_{π∈C_{t}}⟨u|(sgn π)π^{−1} {t}⟩ (since ⟨ | ⟩ is S_{n}-invariant.)

= ∑_{π∈C_{t}}⟨u|(sgn π^{−1})π^{−1} {t}⟩

= ⟨u| ∑_{π∈C_{t}}(sgn π)π {t}⟩

= ⟨u|e_{t}⟩.

Since the e_{t} generate M′_{μ}, it must be that U ⊂ M′_{μ}^{⊥}.

Theorem 3.5. M′_{μ} is irreducible.

Proof. By the previous theorem, the only submodule of M_{μ} contained in M′_{μ} is M′_{μ}; M′_{μ} has no nontrivial submodules, so is irreducible.

Note that M_{μ} is also derived from inducing the trivial representation on S_{μ_{1}} × ⋯ × S_{μ_{n}}. So this theorem tells us how to get S_{n}’s irreducible characters from characters on lower symmetric groups, a kind of induction.

4. An example

While the theorem of Section 3 tells us a lot about the irreducibles of S_{n}, it doesn’t give the powerful theorem we used in Section 2. Let’s return to this formula.

∑_{π} sgn(π)R(μ′ − π) is irreducible.

Since we have actually classified the irreducibles of S_{n}, we can rewrite this:

M′_{μ} = ∑_{π} sgn(π)M_{μ′−π}.

Or, equivalently:

M′_{μ} ⊕ ⨁_{π∈S_{n}\A_{n}} M_{μ′−π} ≅ ⨁_{π∈A_{n}} M_{μ′−π}.

I won’t prove this formula—the proof involves a lot of tricky combinatorics. However, I will show how this works in the elementary case of μ = (1, 1). This demonstrates the key algebraic concepts.

Theorem 4.1.

M′_{{(1,1)}} = ∑_{π∈S_{2}} sgn(π)M_{{(1,1)+(1,2)−π)}}.

Proof. We wish to show

M′_{{(1,1)}} ⊕ M_{{(0,2)}} ≅ M_{{(1,1)}}.

M′_{{(1,1)}} is generated by the polytabloids {1} {2} − {2} {1} and {2} {1} − {1} {2}. M_{{(0,2)}} is generated by the tabloid {1 2}. M_{{(1,1)}} is generated by the tabloids {1} {2} and {2} {1}. For the isomorphism, send

{1} {2} to {1} {2} − {2} {1} + {1 2}, and

{2} {1} to {2} {1} − {1} {2} + {1 2}.

This respects all the relations on S_{n}.

The point of this example is when we pass from M_{(1,1)} to S_{(1,1)}, we lose the distinction between {1}{2} and {2}{1}—in M′_{(1,1)}, they are linear multiples. By adding the {1 2} as a direct summand, we are able to fix sign for these again. At the same time, since {1 2}, as a tabloid, is fixed by all the permutations fixing {1}{2} and {2}{1}, adding it doesn’t interfere with any of the algebraic relations between S_{n} and S_{(1,1)}.

References

1. G. D. James, The representation theory of the symmetric group, Lecture Notes in Mathematics, no. 682, Springer-Verlag, New York, US, 1978.

2. R. Steinberg, A geometric approach to the representations of the full linear group over a galois field, Transactions of the American Mathematical Society 71 (1951), no. 2, 274.