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$G$ is defined by the relations $x^m=y^n=1,xy=yx^k$. For which $m,n,k$ does this give a nonabelian group order $mn$?

I started playing around with this using GAP, starting with the case where $m=13$.

For $n=3$ only $k=3,9$ worked, and they generated isomorphic groups, similarly for $n=9$. For $n=5,7,11$ no value of $k$ worked. For $n=2$, $k=12$ worked. For $n=4$, $k=5,8$ worked and generated isomorphic groups. $k=12$ also worked but generated a non-isomorphic group. Similarly for $n=8$. For $n=6$, $k=3,4,9,10,12$ worked. The groups for $k=3,9$ were isomorphic. and those for $k=4,10$ were isomorphic.

For $n=12$, $k=2,\dots,12$ all worked. Those for $k=2,6,7,11$ were isomorphic, those for $k=3,9$ were isomorphic, those for $k=4,10$ were isomorphic, and those for $k=5,8$ were isomorphic.

I am having difficulty figuring out what is going on. Is all this well-known material? Is there a complete or partial answer to the opening question (for which $m,n,k$ do we get a non-abelian group order $mn$)? If so, where can I find it?

Also how does one do this stuff by hand? I tried a few cases and found it quite tricky to prove that $x=1$, as often happened for the $k$ that failed.

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  • $\begingroup$ Note that if $n=2$ and $k=m-1$, you get a dihedral group. Not very complete, I know :) $\endgroup$ – Patrick Stevens Feb 21 '18 at 19:25
  • $\begingroup$ Why not have a look at $y^{-n}xy^n$? $\endgroup$ – ancientmathematician Feb 21 '18 at 20:58
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    $\begingroup$ The first thing to note is that the relation can be rewritten as $y^{-1}xy = x^k$. So the requirement is that conjugation by $y$ should give an automorphism of $\langle x\rangle$ of order dividing $k$. Since the automorphism group of $\langle x\rangle$ has order $\varphi(m)$, this immediately gives some restrictions. On the other hand, once these restrictions are satisfied, one can construct the desired group as a semidirect product. $\endgroup$ – Tobias Kildetoft Feb 22 '18 at 7:01
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    $\begingroup$ @ancientmathematician Yes, that is an excellent point. So the condition is $k^n=1\bmod m$. Many thanks. $\endgroup$ – almagest Feb 23 '18 at 10:53
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    $\begingroup$ I suppose to be complete one also needs to see that with this necessary condition one can actually find a non-abelian group meeting this spec. I usually look in the group of all $t\mapsto at+b$, $b\in\mathbb{Z}_m$, $a\in\mathbb{Z}^{*}_m$ for the appropriate subgroup. $\endgroup$ – ancientmathematician Feb 23 '18 at 11:50

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