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 $Q=C_8$ we note 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 taken $b\in N$ an element of order $3$, we have that the conjugation $q\mapsto q^b$ is a non trivial automorphism of $Q$. How can we show that this automorphism has order $3$?

Direct computations seems don't lead anywhere.

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

  • $\begingroup$ Does this follow from Lagrange's theorem since $b$ is a nontrivial element of $N/Q$? $\endgroup$ – George Shakan May 30 '14 at 16:54
  • $\begingroup$ Since $b$ has order $3$ it must induce an automorphism of order $3$. What's the problem? $\endgroup$ – Derek Holt May 30 '14 at 19:15
  • $\begingroup$ I can't understand why $|b|=3$ implies that $\psi:q\mapsto q^b$ has order $3$: $\psi^3(q)=(q^b)^3=(q^3)^b$ but I can't see how this implies that $\psi^3=id_Q$... unless we write $(q^b)^3=q^{(b^3)}$, but it seems senseless $\endgroup$ – Joe May 30 '14 at 20:18
  • $\begingroup$ Yes, $\psi(q) = q^b$ implies $\psi^3(q) = q^{b^3} = q$. $\endgroup$ – Derek Holt May 31 '14 at 8:26

Another idea:


are all elements of order four...


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