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In the exercises on direct product of groups of Dummit & Foote, I proved that the symmetric group $S_n$ is isomorphic to a subgroup of $GL_n(K)$, called the permutation matrices with one 1 in each row and each column.

My question is how can I find the normalizer of this subgroup in $GL_n(K)$?

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Try first to find the centralizer. Then use that $S_n$ is complete for $n\neq 2, 6$. –  user26857 Feb 24 '13 at 16:02
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Further hint 1: The fact that $S_n$ is complete implies that its normalizer is the subgroup generated by $S_n$ and its centralizer. –  user3533 Feb 24 '13 at 16:31
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Further hint 2: Each matrix in the centralizer of $S_n$ has exactly 2 distinct values as entries (but it also has a very specific structure). –  user3533 Feb 24 '13 at 16:34

1 Answer 1

up vote 4 down vote accepted
+50

Edit: I've revised the answer to make it more elementary, and to fix the error YACP pointed out (thank you).

Suppose $X\in N_{GL_n(K)}(S_n)$. Then for every permutation matrix $P\in S_n$ we have $XPX^{-1}\in S_n$, so conjugation by $X$ is an automorphism of $S_n$. If $n\ne 2, 6$, then as YACP noted it must be an inner automorphism, i.e. we have some $P'\in S_n$ such that for every $P\in S_n$, $XPX^{-1}=P'P{P'}^{-1}$. Thus $(X^{-1}P')P(X^{-1}P')^{-1}=P$, so $X^{-1}P'\in C_{GL_n(K)}(S_n)$ (the centralizer of $S_n$). Thus $X\in C_{GL_n(K)}(S_n)\cdot S_n$, so all we have to do is find $C_{GL_n(K)}(S_n)$, as $C_{GL_n(K)}(S_n)\cdot S_n\subseteq N_{GL_n(K)}(S_n)$ holds trivially.

Let $\mathcal C$ denote of all matrices (including non-invertible ones) $X$ such that $PXP^{-1}=X$ for all $P\in S_n$. Note that conjugation is linear, i.e. $A(X+Y)A^{-1}=AXA^{-1}+AYA^{-1}$ for any $A,X,Y\in M_{n\times n}(K)$, so $\mathcal C$ is closed under addition. Conjugation also respects scalar multiplication, i.e. $AcXA^{-1}=cAXA^{-1}$, so $\mathcal C$ is closed under scalar multiplication. Recall that $M_{n\times n}(K)$ is a vector space over $K$, so this makes $\mathcal C$ a subspace of $M_{n\times n}$. The use of $\mathcal C$ is that $C_{GL_n(K)}(S_n)=\mathcal C\cap GL_n(K)$, yet unlike $C_{GL_n(K)}(S_n)$ it is a vector subspace, and vector subspaces are nice to work with.

It is easy to see that $\mathcal C$ contains diagonal matrices $D$ with constant diagonal, as well as all matrices $M$ such that the entries $m_{ij}$ are the same for all $i,j$. Since $\mathcal C$ is a vector subspace, this means it contains all sums of these matrices as well. We want to show that every matrix in $\mathcal C$ can be written as $D+M$ where $D$ and $M$ are as above. If $X\in \mathcal C$ then we can subtract a diagonal matrix $D$ and a matrix $M$ of the second kind to get the upper left and right entries to be $0$: $$X-D-M=\begin{pmatrix} 0 & x_{12} & \cdots & x_{1n-1} & 0\\ x_{21} & x_{22} & \cdots & x_{2n-1} & x_{2n}\\ \vdots & \vdots & \ddots & \vdots & \vdots\\ x_{n1} & x_{n2} & \cdots & x_{nn-1} & x_{nn}\\ \end{pmatrix}$$ Call this matrix $X'$; we wish to show $X'=0$. Exchanging the second and last column must be the same as exchanging the second and last row, and since the first action switches $x_{12}$ and $x_{1n}$ while the second leaves the first row unchanged we have $x_{12}=x_{1n}=0$. Continuing in this manner we see that the whole first row is $0$. Exchanging the first and second row is the same as exchanging the first and second column, so the whole second row must be $0$ as well. Continuing in this manner we get that $X'=0$ as desired. Thus $\mathcal C$ is the set of matrices of the form $D+M$, i.e. with $a$ on the diagonal and $b$ off the diagonal.

$C_{GL_n(K)}(S_n)$ is the set of such matrices with nonzero determinant. Let $X\in \mathcal C$ have entries $a$ on the diagonal and $b$ off it. Clearly if $a=b$ then the determinant is $0$, so suppose $a\ne b$. Then we can write $X=(a-b)(I_n+cr)$ where $c$ is a column consisting entirely of $1$'s and $r$ is a row consisting entirely of entries $\frac{b}{a-b}$. By Sylvester's Determinant Theorem the determinant of this is $(a-b)^n(1+rc)$, and $rc=\frac{nb}{a-b}$, which gives us $\det(X)=(a-b)^{n-1}(a-b+nb)$. Thus for any $X\in \mathcal C$, $\det(X)=0$ iff either $a=b$ or $a=(1-n)b$.

Putting this all together, we get that $$N_{GL_n(K)}(S_n)=\left\{\begin{pmatrix} a & b & \cdots & b \\ b & a & \ddots &\vdots \\ \vdots &\ddots & \ddots & b\\ b & \cdots & b & a\\ \end{pmatrix}P: a\neq b, a\neq (1-n)b, P\in S_n \right\}$$

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Thanks for the answer. But, I completely lost you at $S_n$ forms a $K-$ module $M$. I don't know what a module is. This is a question in group theory Also, how can you see that M contains diagonal matrices with constant diagonal as well as all matrices with all entries equal? –  ramanujan_dirac Feb 26 '13 at 18:45
    
@ramanujan_dirac I tried to make the answer more elementary. Let me know if there's something that's still unclear. –  Alex Becker Mar 5 '13 at 1:34
    
I believe the approach of your first answer is the correct way to do this. The question posed by the OP is not in the exercises of Dummit and Foote (at least not where they were working), and thus I believe the onus is on the OP to interpret your first, superior, answer. –  user641 Mar 5 '13 at 2:38
    
@SteveD I haven't lost any the content of the first answer, only the terms. The generalization of the concepts is obvious to those familiar with modules. –  Alex Becker Mar 5 '13 at 2:47
    
@AlexBecker: Sorry for not responding to your answer earlier. I was a bit busy. Brilliant answer! –  ramanujan_dirac Mar 7 '13 at 23:40

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