# Exploring $GL_2(\mathbb F_3)$

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

• 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 $Q$!). – Derek Holt May 30 '14 at 8:25
• 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 – Joe May 30 '14 at 13:14

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.
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.)
• 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, – Matt E May 30 '14 at 0:28
• Ok ok, $PGL_2(\mathbb F_3)\simeq S_4$. Thank you Matt. – Joe May 30 '14 at 0:37