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Let $G$ be a finite group, and $g,h\in G$ be such that $G = \langle g,h\rangle = \langle g,hgh^{-1}\rangle$ - that is, $g,h$ generate $G$, and so does $g,hgh^{-1}$.

Let $n := |G|$, $r := |g|$, and $e := |ghg^{-1}h^{-1}|$ ("$|.|$" denotes "order")

I was doing some computations with covers of curves, and it seems that my computations give the following relation: $$e \ge \frac{n}{2+n-\frac{2n}{r}}$$ The cleanest special case is if $r = 2$, in which case one gets $e\ge \frac{n}{2}$ (in which case we must have $e = \frac{n}{2}$ barring the trivial case when $G$ is cyclic, so that the commutator subgroup is cyclic of index 2)

Firstly, is this true? (I've checked it to be true for the smallest 10 or so finite simple groups, where it seems to hold).

If so, this is somewhat curious to me. Is this related to any known results?

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    $\begingroup$ Wait... if $e$ is the order of $[g,h]$, then you necessarily have $e\leq n/2$: the order cannot be equal to $n$, since then $G$ would be cyclic and $ghg^{-1}h^{-1}$ would be trivial; and so must be a proper divisor of $n$, hence at most equal to $n/2$. Are you actually asserting that when $g$ is of order $2$, then the order $e$ of $[g,h]$ must be equal to $n/2$, and so that the commutator subgroup must be either of index $2$ or equal to all of $G$? $\endgroup$
    – Arturo Magidin
    Commented Jun 21, 2017 at 3:55
  • $\begingroup$ @ArturoMagidin Yes I suppose that is what I claim. Is that obviously wrong? The first example I checked was the case $G = D_{2k}$, where $g$ is a reflection, and $h$ is either a primitive rotation or another reflection which generates with $g$. In either case $g,hgh^{-1}$ generate $G$ iff $k$ is odd, and in that case $e$ is indeed equal to $k$, which is also the order of the commutator subgroup. $\endgroup$
    – stupid_question_bot
    Commented Jun 21, 2017 at 4:09
  • $\begingroup$ Seems like a strong condition, and ex nihilo I wouldn't expect it to be true; but the condition you have on the generators may be stronger than it seems at first glance... $\endgroup$
    – Arturo Magidin
    Commented Jun 21, 2017 at 4:11
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    $\begingroup$ For any real numbers, if $e,r \geq 3$ and $n > 0$, then $e \geq \frac{n}{2+n-\frac{2}{r}}$. $\endgroup$
    – Zach Teitler
    Commented Jun 21, 2017 at 5:16
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    $\begingroup$ There was a typo in my comment. Sorry. I meant: If $e,r \geq 3$ and $n>0$, then $e \geq \frac{n}{2+n-\frac{2n}{r}}$. And it also holds automatically if $e=2$ and $r \geq 4$. I guess the $r=2$ case, and maybe the $e=2,r=3$ case, could still be interesting! $\endgroup$
    – Zach Teitler
    Commented Jun 21, 2017 at 6:01

1 Answer 1

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Denote $f=hgh^{-1}$.

1) If $r\geqslant 4$, then $n\geqslant 4$ and either $e=1$, $r=n$ or $e\geqslant 2>\frac{n}{2+n/2}\geqslant \frac{n}{2+n-2n/r}$.

2) If $r=3$, then either $e\geqslant 3\geqslant \frac{n}{2+n/3}=\frac{n}{2+n-2n/3}$ or $e=2$, then we have $f^3=g^3=1$, $(fg)^2=1$. It follows that the group generated by $f,g$ contains at most 12 elements, since it is a factor of the group $\langle f,g|f^3=g^3=1,(fg)^2=1\rangle=A_4$. But for $n\leqslant 12$ we have $e=2\geqslant \frac{n}{2+n/3}$.

  1. $r=2$. Then $n$ does not exceed the order of the group generated by relations $f^2=g^2=1$, $(fg)^e=1$. This group contains at most $2e$ elements as desired. Indeed, all elements have the form either $(fg)^k$, or $(fg)^kf$, where $k=0,1,\dots,e-1$.
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