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$\DeclareMathOperator\Aut{Aut}\DeclareMathOperator\PSL{PSL}$Is every finite simple group contained in a group of the form $\PSL(n,p)$ for some integer $n\ge 1$ and prime $p$?

More generally, I'd like to understand how general the subgroups of $\Aut(A)$ for $A$ finite abelian can be. The relation with the titular question is that the composition factors of $\Aut(A)$ are all either abelian or of the form $\PSL(n,p)$.

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    $\begingroup$ Every finite group embeds into a finite symmetric group (a group acts on itself by left multiplication), so it is enough to answer for $S_n$, no? $\endgroup$
    – Sam Hopkins
    Commented Jun 21 at 22:15
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    $\begingroup$ @SamHopkins, re, in fact in $\operatorname A_n$, since they're all perfect (except for the cyclic groups of order $p$, which can easily be handled separately). $\endgroup$
    – LSpice
    Commented Jun 21 at 22:17
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    $\begingroup$ This MSE question discusses embeddings of all finite groups in many classes of finite groups, but doesn't directly mention projective special linear groups (probably not hard to achieve though): math.stackexchange.com/questions/27132 $\endgroup$
    – Sam Hopkins
    Commented Jun 21 at 22:18
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    $\begingroup$ @SamHopkins, re, for a simple group that is not cyclic (and so is perfect), an embedding in $\operatorname{GL}(n, p)$ lands in $\operatorname{SL}(n, p)$; and then the pre-image of the scalar matrices is a cyclic, normal, hence trivial, subgroup, so that the composition with the projection to $\operatorname{PSL}(n, p)$ remains an embedding. (I assume $\operatorname{PSL}(n, p)$ always means $\operatorname{SL}(n, p)$ modulo its scalar matrices, hence, say, equals $\operatorname{SL}(n, p)$ if $n$ equals $p$.) $\endgroup$
    – LSpice
    Commented Jun 21 at 22:35
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    $\begingroup$ Every finite group of order $n$ embeds into $S_n$, which embeds into $\mathrm{Alt}_{n+2}$, which in turn embeds (in the obvious way) into $\mathrm{PSL}_{n+2}(p)$ for every prime $p$. $\endgroup$
    – YCor
    Commented Jun 21 at 23:16

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This has really been answered in comments, but it is possible to go a bit further. Every finite group (simple or not) of order $n \geq 3$ embeds as a subgroup of the alternating group $A_{n+2}$ (embed it via Cayley into the obviously labelled copy of $S_{n}$ inside $S_{n+2}$, and adjoin the transposition $(n+1\ \ n+2)$ to each odd permutation arising). Now for any prime $p$, the alternating group $A_{n+2}$ embeds naturally in ${\rm GL}(n+2,p)$. But by the simplicity of $A_{n+2},$ all permutation matrices appearing must have determinant one, and it is also clear that no non-identity permutation matrix is scalar, so that $A_{n+2}$ is isomorphic to a subgroup of ${\rm PSL}(n+2,p)$. Hence every finite group of order $n \geq 3$ is isomorphic to a subgroup of ${\rm PSL}(n+2,p)$ for every prime $p$.

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  • $\begingroup$ Why $n \ge 3$? And more characters. $\endgroup$
    – LSpice
    Commented Jun 21 at 23:31
  • $\begingroup$ @LSpice: I nearly added that $n \geq 3$ was just for ease of exposition, using the simplicity of $A_{n+2}$. Clearly groups of order $2$ embed into ${\rm PSL}(4,p)$ by inspection, but the fact that $A_{4}$ is not simple would necessitate rewording the proof. $\endgroup$
    – Geoff Robinson
    Commented Jun 21 at 23:39
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    $\begingroup$ You don't need simplicity of the alternating groups in any case. The sign of a permutation is the determinant of its permutation matrix. So the even permutations of N letters all map into the special linear group of order N. $\endgroup$
    – Noam D. Elkies
    Commented Jun 22 at 0:55
  • $\begingroup$ A simple group of order $n\ge 3$ embeds into $\mathrm{Alt}_n$ (this holds for every group of order not $2$ mod $4$), which in turn embeds into $\mathrm{PSL}_n(p)$ for every $p$. $\endgroup$
    – YCor
    Commented Jun 22 at 7:57
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    $\begingroup$ Or note that $A_n$ is the commutator of $S_n$, so ends up in the commutator of $GL(n,p)$, which is contained in $SL(n,p)$. Since $Z(S_n)=1$, it must miss all scalar matrices. $\endgroup$
    – Steve D
    Commented Jun 22 at 13:17

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