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 8×20+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?
Proof that in a simple group of order 168 the intersection of the normalizers of two distinct 7-sylows is of order 3
abstract-algebragroup-theoryproof-verificationsimple-groupssylow-theory
Best Answer
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 $X$.
We can then proceed and show that $|N_G(P)\cap N_G(P')|=3$ for all $P,P'\in X$, $P\neq P'$.
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.