Proof that in a simple group of order 168 the intersection of the normalizers of two distinct 7-sylows is of order 3 Given a simple group $G$ of order 168=$2^3$x3x7 show that the intersection of the normalizers of two distinct 7-sylow subgroups is isomorphic to $\Bbb Z_3$ (or it has order 3). I want to make sure this proof is correct. By Sylow's third theorem and the fact that $G$ is simple the number of 7-sylow subgroups is 8 and I shall denote the set of the the subgroups with $S$. By letting $G$ act on $S$ through conjugation and the fact that it has a single orbit by Sylow's 2nd theorem we get that the normalizer of each 7-sylow groups $P$ is of order 21. So the order of the normalizers of two distinct elements $P$ and $Q$ of $S$ can be 21,7,3, or 1. If it is either 7 or 21 this would mean where p is an element of $P$ such that $P$=$p^{-1}Qp$ which means $Q$=$pPp^{-1}$=$P$ contradicting the fact they are distinct. If the intersection was trivial then one would have 8 subgroups of order 21 with trivial intersection so there would be 8x20+1=161 elements of order not divisible by 2 so there can only be a single 2-sylow which contradicts simplicity of $G$. Is this proof correct?
 A: Your argument would be fine, if your goal were to show that $N_G(P)\cap N_G(P')$ must have order $3$ for at least some pair of Sylow $7$-subgroups $P,P'$. However, the way I read the task is that you are to prove that $|N_G(P)\cap N_G(P')|=3$ for any pair of distinct Sylow $7$-subgroups.
More about that later. Revisiting the argument and setting up the scene for the stronger claim.
A possibly simpler way to get to the point you reached yourself could go as follows. Let $X$ be the set of Sylow $7$-subgroups. You correctly deduced that $|X|=8$ and that hence $|N_G(P)|=21$ for all $P\in X$. Therefore, by Cauchy, there is an element $z$ of order three in $N_G(P)$.
Consider the conjugation action of $H=\langle z\rangle$ on $X$.


*

*The orbits of $H$ on $X$ have sizes $1$ or $3$.

*Because $|X|=8\equiv2\pmod3$, there must be at least two orbits of size $1$.

*Clearly $\{P\}$ is an orbit of size one. If $\{Q\}$ is another, both $P$ and $Q$ are normalized by $H$. In particular $H\le N_G(P)\cap N_G(Q)$.

*Because $N_G(P)$ and $N_G(Q)$ cannot share elements of order $7$, their intersection cannot have order $>3$, so $H=N_G(P)\cap N_G(Q)$.


We can then proceed and show that $|N_G(P)\cap N_G(P')|=3$ for all $P,P'\in X$, $P\neq P'$.


*

*We saw above (you showed this in a different way) that there is another Sylow $7$, $Q\in X$ such that $N_G(P)\cap N_G(Q)$ has order three.

*Let $x$ be a generator of $P$. We know that $x$ does not normalize any Sylow $7$-subgroups other than $P$.

*Clearly 
$$x(N_G(P)\cap N_G(Q))x^{-1}=N_G(xPx^{-1})\cap N_G(xQx^{-1})=N_G(P)\cap N_G(xQx^{-1}),$$
so we see that $N_G(xQx^{-1})$ also intersects $N_G(P)$ in a subgroup of order three (that must be $xHx^{-1}$).

*Repeating the above with powers of $x$ we see that the same holds for every $P'\in X$ that belongs to the $P$-orbit of $Q$.

*But the $P$-orbit of $Q$ in $X$ must have seven elements. Hence it contains all the Sylow $7$-subgroups other than $P$ itself.



At this point we have proven that $N_G(P)$ intersects the normalizers of all the other Sylow $7$-groups in a subgroup of order three. Because we started with an arbitrary $P\in X$, the claim holds for all pairs of Sylow $7$s.

The last step would also follow from the fact that the conjugation action of $G$ on $X$ is transitive.

May be many, if not all, of the steps I wanted to add were obvious to you. I just think that in a first course on this theme you would be expected to include them. Nothing deep going on there.
