Semidirect products and finding normal subgroups

Question

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?

Answer 1

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.

Answer 2

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.