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For example, I want to understand what are different $S_5 \rtimes \langle c\rangle_2$ products.

$\mathrm{Aut}(S_5)=\mathrm{Inn}(S_5)\simeq S_5$, so we can have direct product or $\psi: с \rightarrow \tau \in S_5$, such that $o(\tau ) = 2$.

I know that:

Let $N,H$ be groups, $ϕ:H\to\mathrm{Aut}(N)$ be a homomorphism, $\psi\in \mathrm{Aut}(N)$. Then $N \rtimes_{\phi}H\cong N\rtimes_{\psi\circ\phi}H$

So there will be at most one semidirect product. And we can assume that $\psi (c)=(12)$.

But isn't it the same as direct product? Is it true that if $\phi:H\to\mathrm{Inn}(N)$, than $N \rtimes_{\phi}H\cong N\times H$?

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    $\begingroup$ You might try proving the following: if $K\le G$ is a normal subgroup that is complete, then $G\cong K\times C_G(K)$. Now complete means that every automorphism is inner, and the center is trivial. So $S_n$ for $n\neq 2,6$ is complete. Thus if $G=K\rtimes H$ with $K\cong S_5$, then we have $C_G(K)\cong G/K\cong H$, and $G=K\times C_G(K)\cong K\times H$. $\endgroup$
    – user641
    Commented Feb 10, 2012 at 8:34
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    $\begingroup$ As for the question in your last paragraph, $D_8\rtimes Inn(D_8)$ is not a direct product. $\endgroup$
    – user641
    Commented Feb 10, 2012 at 8:44
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    $\begingroup$ Think about how such an element acts on $K$... $\endgroup$
    – user641
    Commented Feb 10, 2012 at 15:40
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    $\begingroup$ @grozhd: Note: use \langle instead of < and \rangle instead of > for delimiters. $\endgroup$ Commented Feb 10, 2012 at 16:42
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    $\begingroup$ Grozhd: someone has to officially answer it. In this case, a very good choice is for you to answer it yourself (noting that Steve gave you the hint). I would up-vote your proof, you could accept it yourself, and earn zillions of rep, since people with the same question will find a nice answer. $\endgroup$ Commented Feb 10, 2012 at 17:46

1 Answer 1

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Idea by @SteveD: if $K≤G$ is a normal subgroup that is complete, then $G≅K×C_G(K)$. Now complete means that every automorphism is inner, and the center is trivial. So $S_n$ for $n≠2,6$ is complete. Thus if $G=K⋊H$ with $K≅S_5$, then we have $C_G(K)≅G/K≅H$, and $G=K×C_G(K)≅K×H$.

Now we will prove that $G≅K×C_G(K)$, if $K≤G$ is a normal subgroup that is complete.

Proof:

1) $C_G(K)$ is normal subgroup:

Let $c\in C_G(K)$, $g\in G$, $k\in K$ then $(gcg^{-1})k= gcg^{-1}kgg^{-1}$, since $K$ is normal $g^{-1}kg\in K$ and $gc(g^{-1}kg)g^{-1}=g(g^{-1}kg)cg^{-1}=k(gcg^{-1})$, so $gcg^{-1}\in C_G(K)$ and $C_G(K)$ is normal.

2) $C_G(K)\cap K=e$:

$K$ is complete, so $Z(K)=e$, now if $a\in C_G(K)\cap K$ then $a\in Z(K)$, so $C_G(K)\cap K=e$.

3) $KC_G(K)=G$:

Let $g\in G$. For $h \in K$ consider $\phi_g: h\rightarrow ghg^{-1}$. $\phi_g \in Aut(K)=Inn(K)$ ($K$ is complete), so there is an element $k\in K$ such that $ghg^{-1}=khk^{-1}$ for every $h\in K$.

So we can write $g=k(k^{-1}g)$. It remains to show that $k^{-1}g$ belongs to $C_G(K)$:

Let $h\in K$ then: $(k^{-1}g)h=k^{-1}(ghg^{-1})g=k^{-1}(khk^{-1})g=h(k^{-1}g)$.

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