Consider a set $G$ of $n \times n$ matrices with entries $\{0,1,-1\}$ that have exactly one non zero entry in each row a column. Show $G$ is a group and that $G$ is the semi direct product of the symmetric group $S_n$ and the group of diagonal matrices with entries in $\{1,-1\}$ (I'll call it $D$).
Well here's the first part:
Let $A,B,C \in G$. Since matrix multiplication is associative it follows $\forall A,B,C \in G$, $A(BC)=(AB)C$.
Since G is the set of $n \times n$ matrices with entries {0,+1,-1} that have exactly 1 non zero entry in each row a column, it follows $I \in G$. $IA=AI=A$ (where I is the identity matrix)
Since each row and column has exactly 1 non zero entry in each row, it follows $\forall A\in G$, $det(A) \neq0$. It follows $ \forall A\in G, \exists A^{-1}$ such that $ AA^{-1}=A^{-1}A=I$ Thus, every element in G is invertible.
Now for the second part: It makes sense that G is the semi direct product of $S_n$ and the group of diagonal matrices with entries in {+1,-1}. $S_n$ acts on D by just permuting is columns..but how can I show that G is the semi direct product?
Thanks!
p.s I see that there are 5 equivalent statements on wikipedia - If I can show just one is true then G is the semi direct product. How do I know which one to show? Is one better than another?
This is what I think is the easiest way:
Show $\phi: G \to S_n$ given by $\phi(A) = B$, where $A = (a_{ij})$ and $B = (b_{ij})$, and $b_{ij} = |a_{ij}|$, is a homomorphism (in other words, take the absolute value of every entry).
We then have that $\text{ker }\phi = D$, and $\phi|_{S_n} = 1_{S_n}$, and we are done.
(This exhibits a right-split of the short exact sequence $1 \to D \to G \to S_n \to 1$)