$3$-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?
 A: Let $G$ be simple group of order $168$; $N_G(H)$ denotes normalizer of $H$ in $G$.
(1) To get $n_3$, first see $n_7$. It should be $1$ or $8$; by simplicity $n_7=8$. 
(2) Let $P$ be a Sylow-$7$ subgroup, then $n_7=[G:N_G(P)]$ i.e. $|N_G(P)|=21$. 
(3) By $n_7=8$, there is homomorphism $G\rightarrow S_8$, hence from $[G,G]=G$ to $[S_8,S_8]=A_8$.
(4) If $N_G(P)$ is abelian, it would be cyclic of order $21$; in $A_8$ no element has order $21$.
(5) So $N_G(P)$ is non-abelian of order $21=7.3$.
[Added: $N_G(P)$ has unique $7$-Sylow subgroup. The number of $3$-Sylow subgroups in $N_G(P)$ is $1$ or $7$; if it is $1$, then $N_G(P)$ will be $Z_7\times Z_3$, abelian, contradiction. So $N_G(P)$ contains exactly seven $3$-Sylow subgroups; call them $H_1,\cdots, H_7$. Notice that $H_i$'s are also $3$-Sylow subgroups of $G$.]
(6) Thus, if $G$ has exactly seven $3$-Sylow  subgroups, they should be  $H_1,\cdots, H_7$; they will be conjugate in $G$. 
(7) So $3$-Sylow subgroups of $G$ are $H_1,\cdots, H_7$; they are conjugate in $G$; since they are contained in $N_G(P)$ the subgroup $\langle H_1,\cdots, H_7\rangle$ should be contained in $N_G(P)$.
But $\{H_1,\cdots, H_7\}$ is a conjugacy class of subgroups in $G$, subgroup generated by them is normal.
Thus $N_G(P)$ contains a normal subgroup of $G$ generated by $H_i$'s; contradiction!
A: 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 $28$.
