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Let $G$ be a finit group and $T$ be a normal subgroup of $G$ such that $PSL(3,4) \unlhd T \leq Aut(PSL(3,4))$ and $|T|=2|PSL(3,4)|$. If $G= T\rtimes C_{11}$, then what we can say about the structure of the group $G$? (Actually I would like $G$ has an element of order 22 and by $C_n$ I mean the cyclic group of order $n$).

Mikasa
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Tina
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  • +1 to an interesting question. Is it possible that $T$ could have an automorphism of order eleven? Feels a bit unlikely given $\mathbb{F}_{4^5}$ is the smallest relevant field with elements of order elevem, but I really don't know this stuff. – Jyrki Lahtonen Aug 27 '13 at 06:29
  • This is nice if we could prove that 11 does not divide $|Aut(T)|$. – Tina Aug 27 '13 at 09:12
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    The order of ${\rm PSL}(3,4)$ is $20160=2^6.3^2.5.7$ and its outer automorphism group has order 12, so there is no automorphism of order 11, and any such semidirect product must be a direct product. – Derek Holt Aug 27 '13 at 09:51
  • If $G=PSL(3,4)\rtimes C_{11}$, then you are right but we have $G=T\rtimes C_{11}$, where $|T|=2|PSL(3,4)|$. – Tina Aug 27 '13 at 10:20
  • Please include context. The previous two questions are just tiny variations on this question: http://math.stackexchange.com/questions/471043/the-structure-of-a-finite-group-of-order-40320/471277#471277 and http://math.stackexchange.com/questions/456491/non-split-extension-of-the-simple-group-l-34 – Jack Schmidt Aug 27 '13 at 11:04
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    For any finite nonabelian simple group $S$ and any group $T$ with $S \unlhd T \le {\rm Aut}(S)$, ${\rm Aut}(T)$ is (isomorphic to) a subgroup of ${\rm Aut}(S)$. – Derek Holt Aug 27 '13 at 11:08

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$G=T\times C$ using the same ideas as before. How does $C$ act on the socle $S$ of $T$? not by an element of order 11, but rather it centralizes it. It centralizes $T/S$ as well. Heck it centralizes $G/T$, so $C$ centralizes a chief series of $G$, and so is contained in the Fitting subgroup of $G$.

Jack Schmidt
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