# The extension of $PSL_2(q)$ by its outer automorphism group

Let $q=p^f$ be a prime power. Is $P\Gamma L_2(q)$, the automorphism group of $PSL_2(q)$, a semidirect product of $PSL_2(q)$ by its outer automorphism group $Z_{\gcd(2,q-1)}\times Z_f$? If it is not in general, then for which $q$ this holds?

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The answer is no for $q=9$, and yes when either $q$ is even, or when $f$ is odd. I am 99% confident that the extension is non-split when $q$ is odd and $f$ is even, but I am not sure how to go about proving that. I have just done some quick computer calculations, and I can confirm that it is non-split for $q= 9,25,49,81,121,625,729$.
So the answer may be that the extension splits iff $\gcd(d,f)=1$, where $d=\gcd(2,q-1)$. This is equivalent to say the outer automorphism group is cyclic. Then I'm wondering if for $G$ a subgruoup of $P\Gamma L_2(q)$ with $G/PSL_2(q)$ cyclic, $G$ is a split extension of $PSL_2(q)$. For example, what about $G=P\Sigma L_2(q)$? –  Binzhou Xia Oct 20 '12 at 10:23
${\rm P \Sigma L}_2(q)$ always splits, because a complement is generated by the field automorphism of ${\rm SL}_2(q)$ that maps every entry in a matrix to its $p$-th power. ${\rm PGL}(2,q)$ also splits. In the case $q=9$, the other extension by a group of order 2, often denoted by $M_{10}$, does not split, and I think the situation is similar for all odd $q$ and $f$ even. –  Derek Holt Oct 20 '12 at 12:47
For other odd $q$ with $f$ even, does $P\Gamma L_2(q)$ have "similar" subgroups like $M_{10}$ in $P\Gamma L_2(9)$? –  Binzhou Xia Oct 20 '12 at 13:07
Yes the corresponding extension would be by the product of the two outer automorphisms of order 2, one from ${\rm PGL}(2,q)$ and the other from ${\rm P \Sigma L}(2,q)$. –  Derek Holt Oct 20 '12 at 19:50
Is this extension always nonsplit? If it is, this also implies $P\Gamma L_2(q)$ is nonsplit for $q$ odd and $f$ even. –  Binzhou Xia Oct 21 '12 at 0:30