# Can we classify all finite 2-generated groups $G$ such that if $x,y$ generate $G$, then so does $x,yxy^{-1}$?

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?

• No, cyclic groups of prime order don't satisfy this (take $x=e$, $y\neq e$) – YCor Oct 30 '17 at 21:58
• 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. – GreginGre Oct 30 '17 at 22:04
• 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). – YCor Oct 30 '17 at 22:10
• 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 $\langle x,y\rangle=G$. – YCor Oct 30 '17 at 23:18
• 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. – YCor Oct 31 '17 at 13:53

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.

• 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. – YCor Oct 31 '17 at 8:30
• @YCor What you need is Gaschutz's Lemma, as mentioned above, which is now "below". – Luc Guyot Oct 31 '17 at 8:53
• You are right! I wrote the general result modulo other comments but forget to mention the use of Gaschutz's lemma. – M. Farrokhi D. G. Oct 31 '17 at 9:21
• 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 $G$. – YCor Oct 31 '17 at 11:19
• 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 $\{x,y\}$. – YCor Oct 31 '17 at 12:09

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).

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.

• There's an endless list of counterexamples. The question is whether there is an example. – YCor Oct 30 '17 at 23:11
• 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. – YCor Oct 30 '17 at 23:25