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?
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 $V_4 \subset A_4$.