3
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There is an exercise in a linear group's book that: prove that every proper subgroup of $PSL(3,3)$ is solvable.

How can I prove that?

the best.

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    What is the order of the group?2012-08-02

1 Answers 1

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Here's how to avoid the classification. Note $| PSL(3,3) | = 2^43^313$.

A quick number-theoretic check (I used Mathematica) shows the only possible number of Sylow 13-groups, $n_{13}$, is $1$, $27$, or $144$. So let $G$ be your proper subgroup of order $2^a3^b13^c$.

If $c=0$, Burnside's theorem shows $G$ is solvable.

If $c=1$, but $n_{13}=1$, then $G$ has a normal Sylow 13-group, and by quotienting and using Burnside's result again, $G$ is solvable.

If $n_{13}=27$, then $b=3$. Let $P$ be a Sylow 13-group; then $N_G(P)$ has index $27$. If $a=0$ ,then $N_G(P)=P$, and Burnside's transfer theorem implies $G$ is solvable. If $a>0$, then $[PSL(3,3):G]\le 8$, contradicting the fact $PSL(3,3)$ is simple.

If $n_{13}=144$, then $a=4$ and $b=2$ (this is where we use that $G$ is a proper subgroup). Thus $N_G(P)=P$, and by Burnside's transfer theorem, $G$ is once again solvable.

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    In the 144 case, $G$ has index 3. :-)2012-08-02