Injectivity of $G\to S_3$

Question

I'm trying to understand a solution given by awwalker in here.

A group $G$ (with identity element $e$) of order $6$ does have three Sylow 2-subgroups, call them $P_1,P_2,P_3$, next define $$\varphi :G \to S_3, \quad \varphi (g)(P_j) = g^{-1}P_jg.$$ It is well-defined, because conjugation preserves "Sylow $p-$subgroupiness" (open to suggestions) and it's also straightforward to check it's a morphism. The trouble is with injectivity. If $\varphi (g) = \mbox{id}$ i.e $g^{-1}P_j g = P_j$, why would this imply $g=e$?

I tried playing around with it by taking $e\neq a\in P_j$, then write $g^{-1}ag = b$, but the most I could derive is $a^{-1}b \in \mbox{Ker}\varphi$.

If it would somehow follow that $g\in P_j$, then $\mbox{Ker}\varphi \subset P_j$ would be forced, making it trivial.

Is the injectivity claim valid to begin with or what's the trick?

Answer 1

The normalizer $N(P_i)$ is itself a subgroup of $G$ that contains $P_i$, so the only two choices for its order are $2$ and $6$. It can't be $6$ because that would make $P_i$ a normal subgroup of $G$. That means if $g \notin P_i$, then $g^{-1}P_ig \neq P_i$. So any element $g$ that stabilizes all three $2$-Sylow subgroups of $G$ must be in their intersection, which is trivial.