Find a non-trivial semidirect decomposition of the following groups Find a non-trivial semidirect decomposition of the groups $S_n$, $n \geq 3$, $D_{2n}$, $n \geq 3$ and $A_4$. Prove that $A_n$, $n \geq 5$ and $Q_8$ have no non-trivial semidirect decompositions.
How do I approach this problem? I have no idea how to find a semidirect decomposition. Is there a way, other than looking at all possible subgroups of the group?
 A: The idea is to find two complementary subgroups of the group. For example,
$S_n$ is a semidirect product of $A_n$ by $\mathbb{Z}/2$ where the complement of $A_n$ is given, e.g., by $H = ⟨(12)⟩$. The dihedral group $D_{2n}$ is a semidirect product of $\mathbb{Z}/n = ⟨s⟩$ by $\mathbb{Z}/2$ where $H = ⟨t⟩$ is the complement of $⟨s⟩$.
The quaternion group $Q_8$ cannot be a semidirect product of $\mathbb{Z}/4$ by $\mathbb{Z}/2$ . In general, it cannot be a non-trivial semidirect product - see here. What do you know about $A_n,n\ge 5$ with respect to normal subgroups ?
A: Clearly $S_n=A_nT$ where $T$ is the subgroup generated by a transposition.
similarly $D_{2n}$ is the inner product of the subgroup $\langle r \rangle$ and $\langle s \rangle$ where $r$ is rotation by $\frac{2\pi}{n}$ degrees and $s$ is any reflection through opposite vertices.
$A_4$ is a little trickier but the normal subgroup you should be looking for is $\{e,(1,2)(3,4),(1,3)(2,4),(1,4)(2,3)\}$
A: A general result you may apply is:
Decomposition in semidirect product:
Let G be a group, H $\triangleleft$ G and K < G such that HK = G and H $\cap$ K = {e}. Then G $\cong$ H $\rtimes_{\varphi}$ K, where $\varphi$: K $\rightarrow$ Aut(H) is an omomorphism defined by $\varphi$(k)=$\varphi_k$ and $\varphi_k$(h)=kh$k^{-1}$.
You can prove it by building an isomorphism F: (H $\rtimes_{\varphi}$ K) $\rightarrow$ G defined by F((h,k))=hk.
At this point you may find the semidirect decomposition by applying the theorem.
For n $\geq$ 3, H = $A_n$ $\triangleleft$ $S_n$ and K = <(12)> < $S_n$ satisfy the hypotheses of the theorem (you may consider <(12)> $\cong$ $\mathbb{Z}/2\mathbb{Z}$).
For n $\geq$ 3, in $D_n$ you consider a rotation r and a reflection s. So H = <r> $\cong$ $\mathbb{Z}/n\mathbb{Z}$ and K = <s> $\cong$ $\mathbb{Z}/2\mathbb{Z}$ give you a decomposition of $D_n$ in semidirect product.
Now let's see a way to decompose $A_4$: you may prove that $S_4$ has just two normal subgroups which are $A_4$ and the Klein group V $\cong$ $\mathbb{Z}/2\mathbb{Z}\times\mathbb{Z}/2\mathbb{Z}$. In $S_4$, V is given (for example) by <(12)(34),(14)(23)>. By considering the conjugacy classes in $S_4$ and using that "H $\triangleleft$ G $\leftrightarrow$ H = $\bigcup_{x\in H}$Cl(x)", you have that V $\triangleleft$ $A_4$. In $A_4$ you also have 3-cycles which aren't in V. You may prove that H = V and K = <(123)> give you a decomposition of $A_4$ in semidirect product (prove that HK gives you all elements in $A_4$ and verify if the omomorphism $\varphi$: <(123)> $\rightarrow$ Aut(V) is well defined).
About $A_n$ for n $\geq$ 5 the following theorem is very useful:
Theorem: For n $\geq$ 5, $A_n$ is simple (i.o.w. $A_n$ has no non-trivial normal subgroups).
By the theorem it follows that $A_n$ (n $\geq$ 5) doesn't have a semidirect decomposition because it has no non-trivial normal subgroups.
The quaternion group $Q_8$ hasn't a semidirect decomposition because each pair of non-trivial normal subgroups has intersection not-trivial {1,-1} (prove it). So you don't have the hypothesis of trivial intersection.
