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Is there any necessary conditions that must be held in order to guarantee the existence of a non-abelian split metacyclic extension? i.e. for which $m,n\in\mathbb{Z}$ there exist a non abelian split extension G $$1\to\mathbb{Z}_m\to G\to\mathbb{Z}_n\to1$$ G must obviously have the form $G=\mathbb{Z}_n\rtimes_\phi\mathbb{Z}_m$, so the question can be formulated as: "For which $m,n\in\mathbb{Z}$ there exist $\phi:\mathbb{Z}_n\to Aut(\mathbb{Z}_m)$ such that $G=\mathbb{Z}_n\rtimes_\phi\mathbb{Z}_m$ is non-abelian?".

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  • $\begingroup$ Judging by exact sequence you would write $G=\mathbb Z_n\ltimes_\phi\mathbb Z_m$. $\endgroup$ – Alex W May 10 '15 at 1:55
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Nontrivial homomorphism $\phi:\mathbb Z_n\to Aut(\mathbb Z_m)$ there exist if and only if $\gcd(n,\varphi(m))\neq 1$, where $\varphi$ - Euler function. Indeed, let $p$ be a prime divisor of $\gcd(n,\varphi(m))$, then $p\shortmid|\mathbb Z_n|$ and $p\shortmid|Aut(\mathbb Z_m)|$, since $|Aut(\mathbb Z_m)|=\varphi(m)$. From $p\shortmid|\mathbb Z_n|$ follow, that there exist $A\leq \mathbb Z_n$ such,that $\mathbb Z_n/A\cong\mathbb Z_p$. From $p\shortmid|Aut(\mathbb Z_m)|$ follow, that there exist $B\leq Aut(\mathbb Z_m)$ of order $p$. Thus $\mathbb Z_n/A\cong B\leq Aut(\mathbb Z_m)$, therefore there exist nontrivial homomorphism $\phi$.

Conversely, suppose that there exist nontrivial homomorphism $\phi:\mathbb{Z}_n\to Aut(\mathbb Z_m)$. Then $1\neq \phi(\mathbb Z_n)\leq Aut(\mathbb Z_m)$, hence $1\neq|\phi(\mathbb Z_n)|\shortmid|Aut(\mathbb Z_m)|=\varphi(m)$. From the other side, $|\phi(\mathbb Z_n)|\shortmid|\mathbb Z_n|=n$. So $1\neq |\phi(\mathbb Z_n)|\shortmid\gcd(n,\varphi(m))$.

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  • $\begingroup$ Thanks. Do you have any reference for this result? Know about any paper of book dealing with these type of non-abelian extensions? $\endgroup$ – Jose Paternina May 10 '15 at 1:49
  • $\begingroup$ @JosePaternina Unfortunately, I do not know anything special about extensions, except what is in the standard textbooks on group theory. I do not know what to advise. $\endgroup$ – Alex W May 10 '15 at 2:09

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