$3$-Sylows of a simple group of $168$ elements I'm trying to find how many $3$-Sylows there are in a simple group of $168$ elements.
Let $n_3$ denote the number of $3$-Sylows.
By the Sylow theorems I know that the possibilities for $n_3$ are $1,4,7,28$. $1$ is not possible since the group is simple. $4$ is also not possible, because then there would be a morphism $ G \rightarrow S_4$ with a nontrivial kernel, which is also impossible. 
I don't know what to do now, can someone please help me?

The group is isomorphic to ${\rm PSL}(2,7)$, which is a doubly-transitive group of degree $8$. Since $168 = 8 \times 7 \times 3$, a $2$-point stabilizer has order $3$.
By considering a diagonal $2 \times 2$ matrix of determinant one over ${\mathbb F}_7$, whose entries are $w$ and $w^{-1}$ with $w$ of order $3$, you can see that the action on the $8$ points of the projective line consists of two cycles of lenth $3$ and two fixed points.
So the number of Sylow $3$-subgroups is the number of (unordered) pairs of points in the projective line, which is $28$.
