Showing a Group $G$ is a Semidirect Product of $S_n$ and the Group of Diagonal Matrices with Entries $±1$. 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$$g^{-1}$ $=$ $I_n$.
From $\underline {Corollary\,3.2.5}$: Suppose $G$ is a group, $N$ and $A$ are subgroups with $N$ normal, $G$ = $NA$ = $AN$, and $A$ $\cap$ $N$ $=$ {e}. Then there is a homomorphism $\alpha$ $:$ $A$ $\to$ $Aut(N)$ such that $G$ is isomorphic to the semidirect product $N$ $\times$ $A$. 
$S_n$ is isomorphic to $SI(n)$ $=$ {permutations of $I_n$} $\in$ $G$. $H$ is the set of all diagonal matrices, both of these are by their definitions subgroups of $G$. By the definitions given: $H$ $\cap$ $SI(n)$ $=$ {$I_n$}, a trivial intersection.
It remains to show that one of $S_n$ or $H$ is normal in $G$. I think this might be a simple task, but I can not see what I am overlooking. 
 A: Define $|(a_{i,j}):=(|a_{i,j}|)$. Show that for matrices with only one nonzero entry per row and column we have $|AB|=|A||B|$. Not only does this show us the existence of inverses in $G$ (for $A\in G$, $|A|$ is a permutation matrix, has an inverse, hence $B:=A|A|^{-1}$ is a matrix with $|B|=I$, hence a diagonal matrix in $G$, which is its own inverse); but it also gives us a homomophism from $G$ to the permutation matrices with kernel the diagonal matrices. From here on, show that $$0\to \{\pm1\}^n\to G\to S_n\to 0 $$
splits.
A: 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).
