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Looking at Sylow questions on $GL_2(\mathbb F_3)$. we have that $Q$ is the unique $2$-Sylow of $N=SL_2(\mathbb F_3)$. $|Q|=8=2^3$ hence by the classification of groups of order $p^3$, we have 5 possibilies for $Q$: $\;C_8,\;C_4\times C_2,\;C_2^3,\;D_4,\;Q_8$. But looking at On $GL_2(\mathbb F_3)$, we see that we have only one element of order $2$ in $N$, hence the same holds for $Q$, then we must exclude $C_4\times C_2,\;C_2^3,\;D_4$.

Hence $Q=C_8$ or $Q=Q_8$. In order to exclude the case $Q=C_8$ my teacher said that even though $Q\unlhd N$, "$Q$ is not centralized by any $3$-Sylow subgroup of $N$" (there are four $3$-Sylow subgroups of $N$, see again Sylow questions on $GL_2(\mathbb F_3)$.), and so $Q$ has an automorphism of order $3$.

I think that by "$Q$ is not centralized by any $3$-Sylow subgroup of $N$", my teacher mean that although $Q\unlhd N$, and thus $Q^g=Q\;\;\forall g\in N$, calling $B$ a $3$-Sylow of $N$, it's not true that $qb=bq\;\;\forall q\in Q,\;\;\forall b\in B$ (first question: how can I see this?), hence we can define a $\psi\in Aut(Q)$ defined by $\psi(q)=q^b$; and the second question is: how can I prove that $\psi$ has order $3$ (the order of an automorphism is defined as the minmum $n\in\mathbb N$ s.t. $\psi^n=id_Q$)? And if $\psi$ is not the right automorphism, (third question) how can I define an automorphism of $Q$ of order $3$?

Proving that there exists an automorphism of order $3$, we can argue as follows: if by contradiction $Q=C_8$, then $Aut(C_8)\simeq U(\mathbb Z_8)$ hence $|Aut(C_8)|=4$, thus it can't contain any automorphism of order $3$. Then I can conclude that $Q=Q_8$.

Thank you all

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    $\begingroup$ If $Q$ centralized a Sylow $3$-subgroup $B$, then that Sylow $3$-subgroup would be normal in $G$, and there would be a unique Sylow $3$-subgroup, which we know is false. So conjugation by an element of order $3$ in $B$ induces an automorphism of $Q$ of order $3$. Since $C_8$ has no automorphism of order $3$, we must have $Q=Q_8$(which is probably why it was named $Q$!). $\endgroup$ – Derek Holt May 30 '14 at 8:25
  • $\begingroup$ Many many thanks Derek! The only thing I can't see is: why has $\psi:q\mapsto q^b$ order $3$? I'm trying again and again, but I can't prove it at all. I'm trying directly: $\psi(q)=q^b$, hence $\psi^3(q)=(q^3)^b=(q^b)^3$, but I'm not able to show that the last one is equal to $q$ $\forall q\in Q$. What to do? Many thanks again $\endgroup$ – Joe May 30 '14 at 13:14
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Have you thought about exploiting the structure of $PGL(2,\mathbb F_3) := GL(2,\mathbb F_3)/(\pm 1)$? This has order $24$, and acts faithfully on the $4$ points of the projective line over $\mathbb F_3$. Its subgroup $PSL(2,\mathbb F_3)$ of index $2$ thus has order $12$. Given what I've just said, you can probably figure out what these groups are.

This will make it pretty straightforward to answer your questions (and any similar ones).


Alternatively, you can just write down some elements in $GL(2,\mathbb F_3)$ of $2$-power order, and an element of order $3$, and see if they commute. (These are just $2\times 2$ matrices, so it's not very hard to compute with them directly.)

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  • $\begingroup$ thanks a lot. The idea is to do the less computation as I can, hence I must exclude the second part of your answer. Looking at the fist part, on the other hand, it seems really interesting and useful, but I can't see how can I use the tools you suggested in order to solve my problems. Can you give me some explicit hints, please? I reasoned about that, but no light bulb turned on :-(... thank you! $\endgroup$ – Joe May 30 '14 at 0:17
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    $\begingroup$ Dear Joe, How many groups of order $24$ do you know? Given that $PGL_2(\mathbb F_3)$ is acting faithfully on a $4$-element set and has order $24$, what group must it be? Regards, $\endgroup$ – Matt E May 30 '14 at 0:28
  • $\begingroup$ Ok ok, $PGL_2(\mathbb F_3)\simeq S_4$. Thank you Matt. $\endgroup$ – Joe May 30 '14 at 0:37
  • $\begingroup$ These small isomorphism between PSL/PGL and Alt/Sym were very helpful to me, but I will caution that in many ways the matrix groups are actually better behaved and easier to understand once you get comfortable using linear algebra in group theory. $\endgroup$ – Jack Schmidt May 30 '14 at 1:40

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