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Given a cyclic group $C_p$ and an abelian (noncyclic) group $K$ with $|K|$ divides $p-1$. Is it always possible to construct a nonabelian group $G \cong C_p\rtimes K$ with $Z(G) \cong C_p$?? If such group can be constructed, How could the homomorphism $\varphi:K \to \operatorname{Aut}(C_p)$ be defined?

It is clear that if $K$ is cyclic then such group can not be constructed, because in this case we get $G/Z(G) \cong K$ which force $G$ to be abelian and hence $G=Z(G)$ which is not possible.

azif00
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Paul
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  • But $G/Z(G) \cong C_n$ will always be cyclic. – the_fox Jan 10 '21 at 17:48
  • I don't know about the center right now, but $\rm{Aut}C_p\cong C_{p-1}$ and the latter always has a copy of $C_n$, when $n|(p-1)$. You can just use the inclusion. –  Jan 10 '21 at 17:48
  • Dear Chris Custer, and the_fox, I have edited the question and your comments are appreciated. – Paul Jan 10 '21 at 19:22
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    I suspect you want genuine semidirect products (that is, products which are not direct). But if $G = Z(G) K$ for some subgroup $K$ of $G$, $K$ will necessarily be normal in $G$. Do you see why? You are asking if it is always possible to construct such a group, but can you produce even one instance where what you want occurs? – the_fox Jan 10 '21 at 20:49
  • If $G=Z(G)K$ then for each $g\in G$ we have $g=zk: z\in Z(G),k\in K$ Therefore, $K^{zk}=K$ and $K$ is normal in $G$. – Paul Jan 10 '21 at 22:45

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