How many $2$-Sylow subgroups does $SL_2(\mathbb{F}_3)$ have?
I wanted to know if my proof is correct -
Since $|SL_2(\mathbb{F}_3)|=24$, we can see that $n_2|3$, and therefore $n_2=1,3$.
Assuming that $n_2=3$, we get that the are $3*8$ elements of order $2$, meaning the whole group.
It can't be, as $3|24=|(SL_2\mathbb{F}_3)|$, and by the first Sylow theorem there should be a subgroup of order $3$.
Therefore, $n_2=1$, and there is only one $2$-Sylow subgroup.
Is my proof correct?
As was pointed out in the comments, you need to try something else. Here's one sketch of a possibility for that something else:
Use the fact that $SL(2,3)$ has a center $Z$ of order $2$ and $SL(2,3)/Z\cong A_4$. Since $A_4$ has only four $2$-elements, $SL(2,3)$ can only have eight, so $n_2=1$. Note that you can also use this isomorphism to show that $n_3=4$.
And if you don't already know that $SL(2,3)$ mod its center yields $A_4$, go ahead and try to prove it. You'll end up understanding the structure of $SL(2,3)$ a lot better. (The main other thing to know about $SL(2,3)$ is that its $2$-Sylow is quaternion--try showing that, too.)
Hint on showing $SL(2,3)/Z\cong A_4$: Remember that $SL(2,3)/Z\cong PSL(2,3)$ which acts faithfully on the projective line over $\Bbb{F}_3$. That line has four points, so $PSL(2,3)$ must be isomorphic to an order $12$ subgroup of $S_4$. How many candidates are there? :)