Let us suppose that $G$ is a simple group of order 90. Show that 2 distinct 3-Sylows cannot contain the same element $g \neq e_G$ where $e_G$ is the identity.
First we can compute the number of 3-Sylows: there are 10 3-Sylows. Let T, S be 2 distinct 3-Sylows. We want to prove that $T \cap S = e_G$
How to do so? Is there a general methodology?
My idea is to consider $g \neq e_G \in T \cap S$ and to show that it is no possible.
We consider the centralizer $Z(g) = \{h \in G, hg=gh\}$ of $g \in G$.
$Z(g)$ is a subgroup of $G$ and contains T,S. $Z(g)$ is of order 18, 45 or 90.
Then we would need to show that $Z(G)$ cannot be of order 18, 45 or 90 for various reasons that make $G$ not simple anymore, but I don't know how to do so.
- If $|Z(g)|=90$ then $Z(g)=G$ and ?
- If $|Z(g)|= 45$ then $Z(g)$ is of index 2 in $G$ and?
- If $|Z(g)|= 18$ then ?
If $|Z(g)|=90$, $Z(g)=G$, so $g$ is central, hence generates a (proper) normal subgroup.
If $|Z(g)|=45$, $Z(g)$ has index $2$, as you noted, and it's well known that subgroups of index $2$ are always normal.
If $|Z(g)|=18$, the group action of $G$ on $G/Z(g)$ by multiplication on cosets gives a homomorphism $G\to\operatorname{Sym}(G/Z(g))\simeq S_5$, since $Z(g)$ has index five. This morphism is not trivial since not every element of $G$ induces the identity permutation. So if $G$ is simple, the map must have trivial kernel, implying $G$ embeds in $S_5$, but $90\nmid 120$.
Added: In more detail, each $x\in G$ defines a permutation $\rho_x$ on the set of $5$ cosets of $Z(g)$ in $G$ by the rule $\rho_x(hZ(g))=xhZ(g)$. The map $\rho\colon G\to\operatorname{Sym}(G/Z(g)):x\mapsto \rho_x$ is a group homomorphism, and $\operatorname{Sym}(G/Z(g))\simeq S_5$ since it is the group of permutations of a set of $5$ elements (the $5$ cosets of $Z(g)$). Since $\rho$ is a group homomorphism, $\ker\rho\unlhd G$. Since $G$ is simple, either $\ker\rho=G$ or $\ker\rho=\{e\}$. The former case means every $\rho_x$ the identity permutation, but this is not true since any element $y$ outside of $Z(g)$ does not induce the identity since $\rho_y(Z(g))=yZ(g)\neq Z(g)$. So $\ker\rho=\{e\}$, meaning $\rho$ is injective, so the image $\rho(G)$ is an isomorphic copy of $G$ in $S_5$, so by LaGrange $|G|$ divides $|S_5|$, contradiction.