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Can we classify finite 2-generated groups $G$ satisfying the following property:

For any pair $x,y$ which generate $G$, the pair $x,yxy^{-1}$ also generates $G$.

By the comments, no nontrivial abelian group can satisfy this property, so I suppose the first question is: Do such groups $G$ exist?

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    $\begingroup$ No, cyclic groups of prime order don't satisfy this (take $x=e$, $y\neq e$) $\endgroup$ – YCor Oct 30 '17 at 21:58
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    $\begingroup$ Extending @YCor remark : If $G$ is abelian and satisfies your property, then for any generating pair $x,y$, we have $G=<x,y>=<x,yxy^{-1}>=<x,x>=<x>$. So $G$ is cyclic. Now take $x=e$ and $y$ a generator of $G$. Then you get $G=\{e\}$. Hence, no nontrivial abelian finite group can satisfy your property. $\endgroup$ – GreginGre Oct 30 '17 at 22:04
  • $\begingroup$ I guess it's easy to prove that $Alt_n$ never satisfies this for any $n\ge 5$. For instance use $x=(123)$ and $y=(12\dots n)$ when $n$ is odd (I'm pretty sure this generates although I haven't fully checked). $\endgroup$ – YCor Oct 30 '17 at 22:10
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    $\begingroup$ Since I took time trying to find an example among the simple groups $G=PSL_2(q)$ (because they have few subgroups), let me mention that none works. Indeed, consider a generator $t$ of $F_q^*$, and $y=\begin{pmatrix} 1 & -1\\ 1& 0\end{pmatrix}$. Denote $M(u,v)=\begin{pmatrix} u & v\\ 0& u^{-1}\end{pmatrix}$ and define $x=M(t,0)$. Then $yxy^{-1}=M(t^{-1},v)$ for some $v\neq 0$, and it follows that $\langle x,yxy^{-1}\rangle$ is the upper triangular group $T$. Since $T$ is maximal and $x\notin T$, it follows $\langle x,y\rangle=G$. $\endgroup$ – YCor Oct 30 '17 at 23:18
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    $\begingroup$ The question is now settled thanks to Guyot and Farrokhi's answers: reduction to simple groups and case of simple groups. Let me mention, on the other hand, that there are (infinite) 2-generated groups with this property. Namely, consider a 2-generated simple group in which every proper subgroup is cyclic. Consider $x,y$ such that $x,yxy^{-1}$ do not generate; so they belong to one cyclic subgroup. So they both commensurate a nontrivial cyclic subgroup $C$; since $C$ has cyclic commensurator, it follows that $x,y$ don't generate. $\endgroup$ – YCor Oct 31 '17 at 13:53
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Theorem The only finite group satisfying the condition is the trivial group.

Proof. Let $G$ be a nontrivial finite group and $S$ be a simple quotient of $G$, which satisfies the condition by Gaschutz's lemma. Then $S$ is non-abelian as mentioned above. If $x\in S$ is any involution, then by well-known result of Guralnick and Kantor in Probalistic generation of finite simple groups, there exists an element $y\in S$ such that $S=\langle x,y\rangle$ while $\langle x,yxy^{-1}\rangle$ is a dihedral group, that is, $S\neq\langle x,yxy^{-1}\rangle$, a contradiction.

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  • $\begingroup$ This settles simple groups, but for the general case I think there is a gap: it's not clear at all that the property passes to quotients: indeed when one gets $x,y$ in the quotient, there is no reason that they can be lifted to generators of the original group. So I first don't see why you can discard a cyclic quotient of prime order, and second even with a simple non-abelian quotient I'm not convinced. $\endgroup$ – YCor Oct 31 '17 at 8:30
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    $\begingroup$ @YCor What you need is Gaschutz's Lemma, as mentioned above, which is now "below". $\endgroup$ – Luc Guyot Oct 31 '17 at 8:53
  • $\begingroup$ You are right! I wrote the general result modulo other comments but forget to mention the use of Gaschutz's lemma. $\endgroup$ – M. Farrokhi D. G. Oct 31 '17 at 9:21
  • $\begingroup$ Great, thanks. Let me state Gaschutz' lemma: let $G$ be a finite group generated by $n$ elements. Then any generating $n$-tuple of any quotient of $G$ can lifted to a generating $n$-tuple of $G$. $\endgroup$ – YCor Oct 31 '17 at 11:19
  • $\begingroup$ Concerning the reference to Guralnick and Kantor, from their theorem I, I understand that it follows: for every non-abelian finite simple group $G$, there exists $s\in G$ such that for every $x\in G\smallsetminus\{1\}$, there exists $t\in G$ such that $\langle x,tst^{-1}\rangle=G$. And thus that every element $x$ of order 2 is part of a generating pair $\{x,y\}$. $\endgroup$ – YCor Oct 31 '17 at 12:09
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This is only a long comment.

Claim. If $G$ is a $2$-generated finite group such that $(x, yxy^{-1})$ generates $G$ whenever $(x, y)$ does, then $G$ is perfect.

Proof. Following this MSE answer, we can assume that the abelianization $G_{ab} \Doteq G/[G, G]$ of $G$ is cyclic. As a result, the group $G$ has a generating pair a component of which lies in $[G, G]$. To see this, apply Nielsen transformations to the image of a generating pair of $G$ in $G_{ab}$. Hence $G = [G, G]$.

As a perfect soluble group is just a trivial group in disguise, there is no non-trivial finite soluble group satisfying OP's condition (which echoes a discussion initiated in the comments).

Side note. Since OP's property is stable under taking quotients (use Gaschutz's Lemma to lift generating pairs), the non-existence of such groups will be established if it is shown that no finite simple groups satisfy OP's property (see YCor's comment for early work in this direction).

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As shown in this MSE answer, the subgroup generated by a conjugacy class in any finite group with a non-cyclic abelianization is proper. So, any such 2-generated group is a counterexample.

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  • $\begingroup$ There's an endless list of counterexamples. The question is whether there is an example. $\endgroup$ – YCor Oct 30 '17 at 23:11
  • $\begingroup$ Btw the fact you mention is completely trivial: the image of a conjugacy class in the abelianization is a singleton. Using maximal subgroups as in the MSE answer you link just makes the argument artificially complicated and restricts the class of groups to which the argument applies. $\endgroup$ – YCor Oct 30 '17 at 23:25

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