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I have a group $G \cong P \rtimes \mathbb{Z}_4$, where $|P| = p^2$ ($p$ is an odd prime). Do I have enough information to determine whether the two possible structures of $P$ both yield unique semidirect products?

If $P$ is cyclic, then $Aut(P)$ is cyclic, and if $P$ is elementary abelian, then $Aut(P) = GL(2,p)$. But I'm stuck here. How do I find the possible homomorphisms $\mathbb{Z}_4 \rightarrow Aut(P)$? And how can I tell if they lead to isomorphic or non-isomorphic groups?

Ultimately I'd like to find the center(s) of the possible groups. I read this post What is the center of a semidirect product?, but I'm not sure if I have enough information. The process seems quite complicated. Also, someone here Semidirect product uniqueness argument for classifying groups of small order mentioned the "normal Sylow trick". What is that? Here $P$ is normal Sylow, hence unique and characteristic, but how does that help?

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Leppala
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    For the case when $P$ is cyclic, there are three isomorphism classes of semidirect products, one of which is the direct product. These correspond to the three possible kernels of your homomorphism $C_4 \to {\rm Aut}(P)$. Try and understand that case first. The elementary abelian case is more complicated, with more possible groups, and it depends on whether $p \equiv 1$ or $-1$ mod $4$. – Derek Holt Jan 21 '15 at 09:54
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    If $Ker(f) = \mathbb{Z}4$, then $f(a) = 1 \ \forall a \in \mathbb{Z}_4$ and this leads to the direct product. Also $Ker(f)$ can be $\mathbb{Z_2}$ or $1$. Doesn't $Ker(f) = 1$ mean that $4 \mid (p-1)$ as $Aut(P) = \mathbb{Z}{p(p-1)}$? – Leppala Jan 21 '15 at 14:38
  • @Leppala: You are on the right track. Also in the cyclic case the solution depends on whether $p = 1$ mod $4$ or not. – j.p. Jan 21 '15 at 15:09
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    Yes, sorry, my previous comment was misleading (i.e. wrong). If $p \equiv 3 \mod 4$ then there is no group with $\ker f = 1$. – Derek Holt Jan 21 '15 at 16:13

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