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Is every finite quasi-simple group generated by 2 elements ?

Recall that G is quasi-simple if G is perfect and G/Z(G) is simple.

Edit: If the answer is yes, do we know of a larger class of finite groups (beyond the finite quasi-simple groups) having the 2-generation property ?

Edited: Yes, there is another class. See my answers in comments.

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    $\begingroup$ I'm a little puzzled why all the downvotes. $\endgroup$ – Todd Trimble Nov 22 '16 at 2:50
  • $\begingroup$ Thank you Todd. Maybe because I found the answer and posted it (as a comment) within an hour of posting the question! It happened because I am in the process of learning finite group Theory> Yesterday I posted in comments that every finite product of non-isomorphic finite simple groups is also generated by 2 elements. This is because every subdirect product subgroup of a finite product G of non-isomorphic finite simple groups coincides with G (an elementary exercise). Probably, it will Not be seen as it should because of the negative votes! $\endgroup$ – Nazih Nahlus Nov 22 '16 at 9:48
  • $\begingroup$ I wrote out a careful answer, including all details to be constructive and try to be helpful ( and I did point out that this sort of argument could be found in many texts). Anyone who knew what the Frattini subgroup was would see immediately what the answer was, even if they had not seen it before. Likewise, it is good that you have learned something in writing out the proof in another case of 2-generation that was noted, but this case would also be clear to experts and is well-known. $\endgroup$ – Geoff Robinson Nov 22 '16 at 15:47
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It turns out the answer is Yes.

This is because G/Z(G) is simple and hence generated by 2 elemnts> Moreover, if 2 elements generate a perfect group modulo its center, then they evidently generate the group since

G = < a, b> Z(G) and G = [G, G]. imply G = < a, b>. That is, G is generated by a and b.

Moreover, G is "quasi-generated" by 1.5 elements in the sense that for all x in G - Z(G), there exists y in G, such that G =< x, y>.

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    $\begingroup$ The easiest thing to note is that if $G = [G,G],$ then $Z(G) \leq \Phi(G),$ the Frattini subgroup ( this is quite similar to the argument you are using). $\endgroup$ – Geoff Robinson Nov 7 '16 at 21:45
  • $\begingroup$ Does this imply my note above that G =< a, b> ? Please clarify. $\endgroup$ – Nazih Nahlus Nov 7 '16 at 22:33
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    $\begingroup$ Well, $Z(G) \leq \Phi(G)$ means precisely that $Z(G)$ is contained in every maximal subgroup of $G$ ( to see this, note that if $M$ is a maximal subgroup which does not contain $Z(G),$ then $G = MZ(G)$ and $G =[G,G] \leq M,$ a contradiction. Hence we must have $\langle a,b \rangle =G,$ for otherwise $\langle a,b \rangle$ is contained in a maximal subgroup $M.$ But then $G = \langle a, b , Z(G) \rangle \leq M, $ a contradiction. Thi sort of standard argument can be found in many texts. $\endgroup$ – Geoff Robinson Nov 7 '16 at 23:05
  • $\begingroup$ Do we know of a larger class of finite groups (beyond the finite quasi-simple groups) having the 2-generation property ? $\endgroup$ – Nazih Nahlus Nov 8 '16 at 8:16
  • $\begingroup$ Regarding my question on the 2-generation, One can show that every finite product of non-isomorphic finite simple groups is also generated by 2 elements. This is because every subdirect product subgroup of a finite product G of non-isomorphic finite simple groups coincides with G (an elemntary exercise). $\endgroup$ – Nazih Nahlus Nov 22 '16 at 0:31

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