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Consider a finite metacyclic group with presentation $$G = \langle x,y \, |\, y^n=1, x^u=y^r, x^{-1}yx = y^k \rangle,$$ where $r \mid n$.

Is it true that if $G$ does not split (i.e. $G$ is a not a semidirect product of cyclic groups), then $\gcd(u,(n/r))\neq 1$?

From direct computations, I have verified that this assertion holds true when $|G| \leq 200$, and furthermore, several other families of metacyclic groups (such as quaternionic groups, dicyclic groups, $p$-groups, etc.) appear to satisfy this property.

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  • $\begingroup$ You have to be a little careful here, because $y$ could have order less than $n$ (depending on $k$), but $y^r$ certainly has order dividing $n/r$, and $x$ has order $u$ modulo $\langle y \rangle$, so if $\gcd(u,n/r) = 1$, then $\langle y^r \rangle$ has a complement, or order $u$, in $\langle x \rangle$, which is also a complement of $\langle y \rangle$ in $G$. So, the answer to your question is yes. $\endgroup$
    – Derek Holt
    Commented Apr 1, 2021 at 8:08
  • $\begingroup$ Could you introduce $k$? you probably want it to be coprime to $n$ (otherwise the resulting group has $y$ of order $<n$). $\endgroup$
    – YCor
    Commented Apr 1, 2021 at 8:17
  • $\begingroup$ But I believe that my argument shows that when $\gcd(u,n/r)=1$, the group is a semidirect product of (possibly trivial) cyclic groups even when $y$ has order less than $n$. And even when $k$ is coprime to $n$, $y$ could have order less than $n$ as in $\langle x,y \mid y^5=x^2=1, x^{-1}yx=y^2 \rangle$. $\endgroup$
    – Derek Holt
    Commented Apr 1, 2021 at 8:19

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