TITLE: More than one $2$-Sylows.
QUESTION [2 upvotes]: I aim to show that $G=PSL(2,4)$ has a subgroup of order 12 ($|G|=60$). I've done it for the case when the number of Sylow $2$-subgroups is more than $1$. I suspect that $|Syl_2(G)| = 1$ is actually not the case, but how to show this?

REPLY [2 votes]: The Sylow 2-subgroup is normal (characteristic, in fact) if it is the only one.  Quotient it out, get group of order 15, of which there is one.  This completely factors $G$, so you should be good to go.

REPLY [0 votes]: $G \cong A_5$. This can be shown a number of different ways. One way is to note that the only subgroup of order 60 in $S_5$ is $A_5$. Then note that $A_5$ has 5 subgroups of order 4. So $|Syl_2(G)|\neq 1$.
{"set_name": "stack_exchange", "score": 2, "question_id": 729371}