It is known that for any finite simple group $G$ there exist two elements $a,b\in G$ such that $a$ and $b$ are conjugates in $G$ and $\langle a, b \rangle=G$. My question: Is it true for any finite non-solvable group?
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4$\begingroup$ Certainly not. E.g., consider $G\times A$ where $G$ is nonsolvable and $A$ is noncyclic abelian. $\endgroup$Emil Jeřábek– Emil Jeřábek2025-10-27 08:05:56 +00:00Commented Oct 27 at 8:05
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8$\begingroup$ There are also anabelian counterexamples (=no abelian composition factors), like $A_5^{20}$. $\endgroup$Sean Eberhard– Sean Eberhard2025-10-27 09:09:16 +00:00Commented Oct 27 at 9:09
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2$\begingroup$ @SeanEberhard That sounds like a fun example. I may try my hand with it as an exercise, but I'm sure I'm not the only reader at MSE who would enjoy having it fleshed out a bit (assuming you find the time to do that, no pressure). $\endgroup$Jyrki Lahtonen– Jyrki Lahtonen2025-10-30 07:20:52 +00:00Commented Oct 30 at 7:20
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3$\begingroup$ @JyrkiLahtonen: this is one of those (in)famous examples where $A_5^{19}$ is 2-generated but $A_5^{20}$ is not. Philip Hall proves this in The Eulerian Functions of a Group. More references can be found in this MO post. $\endgroup$Steve D– Steve D2025-10-30 09:08:13 +00:00Commented Oct 30 at 9:08
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1$\begingroup$ Right, the example wasn't intended to be a deep observation. Actually probably $A_5^n$ for $n = 10$ or something is already a counterexample, even though it is $2$-generated. By Hall, the maximal $n$ such that $A_5^n$ is not a counterexample should be equal the number of pairs of conjugate elements $a, b$ that generate $A_5$, up to $S_5$-conjugacy. $\endgroup$Sean Eberhard– Sean Eberhard2025-10-30 10:01:18 +00:00Commented Oct 30 at 10:01
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