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?



curious relation between orders of generators of a finite group
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 $e=2\geqslant \frac{n}{2+n/3}$.


3. $r=2$. 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 $k=0,1,\dots,e-1$.


