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Which nonabelian finite simple groups contain $PSL(2,q)$ for some $q$?

Obviously $PSL(2,q)$ themselves do. Also, as $PSL(2,4)\cong PSL(2,5)\cong A_5\subset A_n,\; n\geq 5$, alternating (nonabelian simple) groups do as well. I believe I have read somewhere that $PSL(2,q)$ embeds in $PSp(2,q^2)$ as well, though I can't remember the reference so I wouldn't put money on it.

Of the finite simple groups listed by the classification theorem, which others contain a $PSL(2,q), \; q\geq 4$?

I would appreciate a reference since it would be unreasonable to ask for proofs.

ADDENDUM: In response to Derek Holt's comments, let me clarify that I do not need to know which $PSL(2,q)$'s are contained in which simple groups. My purpose is this: I am trying to prove a theorem about simple groups. I have the result for $PSL(2,q)$, for $q\geq 4$, and the family of groups for which it holds is upward closed. This deals with a lot of simple groups (for example all alternating groups, per above), and I am trying to figure out which simple groups I still have to worry about.

Ben Blum-Smith
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    I believe that it would be possible to give a complete answer to this question by going through the lists of maximal subgroups of the simple groups, but very hard work. I would be more willing to help if you asked a more restricted question. Here are a couple quick remarks. It is proved in http://mathoverflow.net/questions/167958 that a simple group contains $A_5 \cong {\rm PSL}(2,5)$ if and only is its order is divisible by $60$, but a result like that will not hold for larger $q$. Also ${\rm PSL}(2,q) = {\rm PSp}(2,q)$ so certainly ${\rm PSp}(2,q^2)$ contains ${\rm PSL}(2,q)$. – Derek Holt Feb 18 '16 at 08:43
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    A few more quick remarks. $A_n$ contains ${\rm PSL}(2,q)$ for all $n \ge q+1$, or $n=q$ for $q=5,7,11$. You can get the answer for the sporadic groups from the ATLAS, but there may still be one or two unknown cases for the Monster. A groups of Lie type in the same characteristic as $q$ will contain $L_2(q)$ whenever the order of the field is a power of $q$. It is more complicated for coprime characteristic. The smallest dimension of an irreducible representation of ${\rm PSL}(2,q)$ is $(q-1)/2$ for $q$ odd and $q-1$ for $q$ even, but – Derek Holt Feb 18 '16 at 11:37
  • @DerekHolt - Tell me if I am getting this right: (1) All sporadic groups contain $A_5=PSL(2,4)$ as their orders are all divisible by 60. (2) All groups of Lie type over $\mathbb{F}_q,; q\geq 4$ will contain $L_2(q) = PSL(2,q)$ (including Chevalley groups, Steinberg groups and Suzuki-Ree groups?) (3) And obviously all alternating groups contain $A_5$. Therefore there is no finite simple group that does not contain $PSL(2,q)$ for some $q\geq 4$! Am I understanding correctly? – Ben Blum-Smith Feb 18 '16 at 18:09
  • I meant "nonabelian finite simple group" in that second-to-last sentence. – Ben Blum-Smith Feb 18 '16 at 18:15
  • @DerekHolt - wait, I suppose I need to worry about groups of Lie type with $q=2,3$? – Ben Blum-Smith Feb 18 '16 at 18:34
  • Not quite! It's not true for ${\rm PSL}(3,3)$ for example. That's the only counterexample I can think of. – Derek Holt Feb 18 '16 at 18:34
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    Sorry, I was wrong about the Suzuki groups. They do not contain any ${\rm PSL}(2,q)$ (this is clear because their orders are not divisible by $3$). The Ree groups of type $^2G_2$ do contain ${\rm PSL}(2,q)$ however. – Derek Holt Feb 18 '16 at 18:46

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I have now convinced myself that all finite nonabelian simple groups apart from ${\rm PSL}(3,3)$ and the Suzuki groups ${\rm Sz}(2^{2n+1})$ contain ${\rm PSL}(2,q)$ for some $q \ge 4$.

I don't feel like writing a detailed proof. It might help you to look at the list of the minimal simple groups (i.e. simple groups with no nonabelian simple group as a proper subgroup) although I guess you would still have to worry about the possibility of a group only having one of the above exceptions as simple subgroups.

Derek Holt
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  • +1 Okay I will try to reproduce this based on your recommended sources. Unfortunately, my university's library doesn't have the ATLAS so it is hard for me to get a hold of, but I'll see if I can make it work. – Ben Blum-Smith Feb 19 '16 at 16:15
  • In fact I think my answer to http://mathoverflow.net/questions/167958 provides enough information for you to reconstruct the proof. In dimensions bigger than $5$ all of the groups of Lie type contain $A_5$, so you only have a few cases to think about. – Derek Holt Feb 19 '16 at 17:21
  • Yes this is just the type of information I needed. For my purposes I now know everything I need to proceed. Thanks again. – Ben Blum-Smith Feb 19 '16 at 20:43