Problem Setup:
We want to show that a group $G$ of order $180$ is not simple. We do this by contradiction.
Thus we first observe that $n_3(G) = 10$, which means that the number of $3-$Sylow subgroups is $10.$ Next, we try to show that the $A\cap B = \{e\}$ where $A\neq B$ and $A,B$ are $3-$Sylow subgroups.
For this, we assume that $g\in A\cap B$ such that $g\neq 1$ then we consider $C(g) =\{h\in G:gh=hg\}.$ Then we can observe that $C(g)$ is subgroup of $G.$
I want to show that $C(g)\supset A$ and $C(g)\supset B$ and that $|C(g)| = 18.$ Apparently this should lead to a contradiction implying that $A\cap B =\{e\}.$
My Attempt:
Here is what I have so far. We know that $A$ and $B$ are conjugates of each other and therefore there exists $y\in G$ such that $A = y^{-1}By.$ Next, if $a\in A$ then in order for it to be in $C(g)$ it must satisfy $ga=ag$ or $g=aga^{-1}$ which implies that $$y^{-1}gy=aga^{-1}\implies gya =yag\implies ya\in C(g).$$ But I can't see how to work backward and show that $A\subset C(g).$
However, if we assume that $C(g)\supset A$ and $C(g)\supset B$ then we observe that $|C(g)| \in \{9, 18,36,45,90,180\}$ since it contains subgroups of order $9$, namely $A$ and $B.$ Now $|C(g)|\neq 9$ since that would mean that $C(g) = A=B$ which is impossible. If $|C(g)| = 180$ then $C(g) = G$ but I am not sure why this would cause a problem.
I think to obtain further reduction I will have to use the fact that $|G|$ divides $[G:Z(g)]!$ since otherwise $Z(g)$ would contain a non-trivial normal subgroup [index theorem] which will be a contradiction. So if we have $|C(g)|\in \{18,36,45,90\}$ then I can show using the index theorem that $|C(g)|=18.$ Although, I don't see how this could lead to a contradiction.
Note:
This is different from the question asked here since I am asking a question that refers to a very specific step in the proof that is used to show that the group is not simple. Furthermore, the way that I have to prove this fact is different from the way used there.
