Consider $G$ to be the set of $n$ $\times$ $n$ matrices with entries in $\{\pm1\}$ that have exactly one nonzero entry in each row and column. These are called signed permutation matrices.

Show that $G$ is a group, and that $G$ is a semi-direct product of $S_n$ and the group of diagonal matrices with entries in $\{\pm1\}$. $S_n$ acts on the group of diagonal matrices by permutation of the diagonal entries.

Here is my solution so far:

I show that $G$ is a group:

1) Associativity of $G$ under the group operation: Matrix multiplication is known to be associative. $G$ associative.

2) Existence of Identity Element: $I_n$ $\in$ $G$, by definition of $G$.

3) Existence of an Inverse Element: Because all g $\in$ G have a non-zero determinant, it is true that $\forall$$g$ $\in$ G, $\exists$$g^{-1}$ $\in$ $G$ such that $g$$g^{-1}$ $=$ $I_n$.

From $\underline {Corollary\,3.2.5}$: Suppose $G$ is a group, $N$ and $A$ are subgroups with $N$ normal, $G$ = $NA$ = $AN$, and $A$ $\cap$ $N$ $=$ {e}. Then there is a homomorphism $\alpha$ $:$ $A$ $\to$ $Aut(N)$ such that $G$ is isomorphic to the semidirect product $N$ $\times$ $A$.

$S_n$ is isomorphic to $SI(n)$ $=$ {permutations of $I_n$} $\in$ $G$. $H$ is the set of all diagonal matrices, both of these are by their definitions subgroups of $G$. By the definitions given: $H$ $\cap$ $SI(n)$ $=$ {$I_n$}, a trivial intersection.

It remains to show that one of $S_n$ or $H$ is normal in $G$. I think this might be a simple task, but I can not see what I am overlooking.

  • $\begingroup$ Your part $3$ is wrong. You only know that an inverse matrix exists, but not (immediately) that this inverse is in $G$. $\endgroup$ – Hagen von Eitzen Jul 8 '15 at 6:33
  • $\begingroup$ I just realized I omitted that part. I have now explicitly shown it, good catch. $\endgroup$ – Alekos Robotis Jul 8 '15 at 6:34
  • $\begingroup$ To see that one is normal, first you ought to see which one, which is already stated in the problem ("$S_n$ acts on..."). From there, it really is just a verification. $\endgroup$ – Tobias Kildetoft Jul 8 '15 at 6:35
  • $\begingroup$ You are proposing that $S_n$ is the Normal subgroup, if I understood correctly. Can you explain to me how you know this? That is, what about "$S_n$ acts on..." tells us this? $\endgroup$ – Alekos Robotis Jul 8 '15 at 6:37
  • $\begingroup$ No, that $S_n$ is the one to act means that the other one is the normal subgroup ($S_n$ will, when viewed as a subgroup, act via conjugation). $\endgroup$ – Tobias Kildetoft Jul 8 '15 at 6:38

Define $|(a_{i,j}):=(|a_{i,j}|)$. Show that for matrices with only one nonzero entry per row and column we have $|AB|=|A||B|$. Not only does this show us the existence of inverses in $G$ (for $A\in G$, $|A|$ is a permutation matrix, has an inverse, hence $B:=A|A|^{-1}$ is a matrix with $|B|=I$, hence a diagonal matrix in $G$, which is its own inverse); but it also gives us a homomophism from $G$ to the permutation matrices with kernel the diagonal matrices. From here on, show that $$0\to \{\pm1\}^n\to G\to S_n\to 0 $$ splits.

  • $\begingroup$ I don't quite understand how the $\left\lvert{(a_{i,j})}\right\rvert$ function is defined, could you explain it in words (my linear algebra is rusty at best). Also, when you say that the mapping at the end splits, what does this mean? $\endgroup$ – Alekos Robotis Jul 8 '15 at 6:56
  • $\begingroup$ Basically, you turn one of your signed permutation matrices into a "standard" permutation matrix (all $+1$'s). $\endgroup$ – David Wheeler Jul 8 '15 at 11:54
  • $\begingroup$ I devised a lemma of sorts for this particular case, which allows for solving without showing that either $S_n$ or $H$ is normal in $G$. $\underline{Lemma}:$ Let $H_1$ and $H_2$ be subgroups of $G$, with $H_1$ $\cap$ $H_2$ $=$ {e}, and $\left\lvert{H_1}\right\vert$$\left\lvert{H_2}\right\rvert$ $=$ $\left\lvert{G}\right\rvert$, the $G$ $=$ $FH$ $=$ $HF$. Clearly this is applicable in this case, because it does not matter which of $S_n$ and $H$ is normal in $G$. $\endgroup$ – Alekos Robotis Jul 8 '15 at 22:20

We can also proceed without starting with an element of $G$ and explicitly writing it as a product of a diagonal matrix and a permutation matrix - instead we can work backwards:

Let $H=\mathrm{diag}(\pm1,\cdots,\pm1)\cong\mathbb{Z}_2^n$. Then $H$ and $S_n$ (defined as permutation matrices) are clearly subsets of $G$, and there is a multiplication map $H\times S_n\to G$. Since $H\cap S_n=\{I\}$ we see this map is injective, and by simply counting we see it's surjective.

Conjugating an element $h\in H$ by a permutation (matrix) $\sigma\in S_n$ simply permutes the diagonal entires of $h$ using $\sigma$, so it is still in $H$. From this follows a bunch of things: $G$ is closed under multiplication and inverses, hence is a group, with subgroup $S_n$ and normal subgroup $H$, and finally that $G=H\rtimes S_n$ (mini-exercise).


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