Semidirect products and finding normal subgroups My question reads: 
Let A, K be subgroups. Group G is called semidirect product of A and K if A  $\trianglelefteq$ G, G=AK and  A$\cap$K = < e >. Show that the groups are the semidirect product of two of its subgroups. 
a) S$_3$
b) D$_4$
c) S$_4$ 
Now I am not sure if this is asking for a proof for each part or to directly pick two subgroups that are normal and then make sure the conditions for semidirect products are met. Also, doesn't this imply I need to show the subgroups I pick are normal? I need help picking these subgroups and from there I think it will be straightforward showing the other conditions are satisfied 
For example for S3 could I pick the whole group itself? 
 A: Per request: In $S_3$ let $A = \langle (1 \ 2 \ 3) \rangle$ and $K = \langle (1 \ 2) \rangle$. Note that the index of $A$ in $S_3$ is $2$ so we are guaranteed that $A$ is normal in $S_3$ (if you like, you can check normality with the definition).
Now our goal is to show that $S_3 = AK$. So let's just compute $AK$. There are $6$ quantities to compute, they are as follows:
\begin{align*}
(1)(1) &= (1)\\
(1)(12) &= (1 \ 2) \\
(1 \ 2 \ 3)(1) &= (1 \ 2 \ 3)\\
(1 \ 2 \ 3)(1 \ 2) &= (1 \ 3)\\
(1 \ 3 \ 2)(1) &= (1 \ 3 \ 2)\\
(1 \ 3 \ 2)(1 \ 2) &= (2 \ 3)
\end{align*}
You see every element of $S_3$ show up there, so $AK = S_3$.
Technically, this computation is unnecessary since $|AK| = \frac{|A||K|}{|A\cap K|}$. But I thought that it would be instructive in this case.
A: In D$_4$, let A=< r>, K={e,s}. Checking that A is normal: by lagrange theorem, have index 2 thus A is normal. 
Now showing D$_4$=AK We compute:
ee=e
es=s
re=r
rs=rs
r$^2$e=r$^2$
r$^2$s=r$^2$s
r$^3$e=r$^3$
r$^3$s=r$^3$s
Then, every element in D$_4$ shows up, so AK=D$_4$.
Then showing there intersection is < e > we can state it by looking at our sets. 
