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Show that $SL_2(F_3)/Z(SL_2(F_3)) \cong A_4$

I know that $|SL_2(F_3)/Z(SL_2(F_3))|= 12$.

Then if the quotient group has a normal subgroup of order $4$ then it is isomorphic to $A_4$.

Suppose that it has a normal subgroup of order $3$ then we need to find a contradiction.

Suppose that $V$ is the normal subgroup of order $3$ in $SL_2(F_3)/Z(SL_2(F_3))$ then $V$ is a normal subgroup of order $6$ in $SL_2(F_3)$. Also we see that $V$ must be congruent to $S_3$.

Then how do we proceed after this? Is there a way to proceed from here. There is another answer on the site, here, which doesn't seem intuitive to me.

Any help would be appreciated.

Shaun
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Antimony
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1 Answers1

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$PSL_2(\Bbb F_3)$ has more than one Sylow $3$-subgroup so that it must be isomorphic to $A_4$ - see the following post:

A group of order $12$ either has a normal $ 3$-Sylow subgroup or is isomorphic to $A_4$

Dietrich Burde
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  • I just wanted to ask is there a way to find all the $4$ sylow $3$ subgroups of $SL_2(F_3)$.I was looking for this but I got only $2$ sylow subgroups. How do I find the rest.Is there any linear algebra trick to find it ? – Antimony May 12 '22 at 02:29
  • You can find the $3$-Sylow subgroups at this site, too. See for example this post and try more conjugations as Hagen says. Of course, it is enough to show that there is more than $1$, which you have done already. – Dietrich Burde May 12 '22 at 07:52