Show that $PSL(2,2)$ and $PSL(2,3)$ are not simple groups.

Question

Show that $PSL(2,2)$ and $PSL(2,3)$ are not simple groups.

My intuition:

Is it enough to say that these two groups are solvable since $PSL(2,2)$ has order $6$ and is isomorphic to the symmetric group $S_3$, and $PSL(2,3)$ has order $12$ and is isomorphic to the alternating group $A_4$. And thus, aren't simple?

There are no simple non-abelian groups of order $6$ and $12$, so you are done. You can use this post for solvability. — Dietrich Burde, Jul 03 '19 at 16:52

score 3 · Accepted Answer · answered Jul 03 '19 at 16:51

3

If you already know that $\text{PSL}(2,2)$ is isomorphic to $S_3$, then you can just specify one non-trivial proper normal subgroup of $S_3$, namely the subgroup $A_3$ generated by a $3$-cycle (by the isomorphism this group will get mapped to a non-trivial proper normal subgroup of $\text{PSL}(2,2)$ then.

Analogously for $A_4$, by considering the Klein-$4$-group $V_4 \subset A_4$.

answered Jul 03 '19 at 16:51

Con

9,119

Thank you. will accept the answer when the cooldown of accepting an answer is done – Ilan Aizelman WS Jul 03 '19 at 16:58
1

Glad that I could help. – Con Jul 03 '19 at 17:00
How to show that $PSL(2,3) \cong A_4$? – Rodolfo Ferreira de Oliveira Mar 13 '22 at 21:35

Show that $PSL(2,2)$ and $PSL(2,3)$ are not simple groups.

1 Answers1