I have this problem in my notes: If $|G|=2^{3}3^{3}11$, then $G$ is not simple
The instructor solved it in a way that I could not follow. The solution I have is attached below. If someone could provide me with another easier solution, I would appreciate it.
Perhaps one can make one simplification:
Assume $G$ is not simple, then $n_{11} = 12$. Let $P \in \text{Syl}_{11}(G)$, then $$ [G:N_G(P)] = 12 $$ The action of $G$ on the left cosets of $N_G(P)$ gives a homomorphism $G\to S_{12}$. Since $G$ is simple, we get an injection
$$ G\hookrightarrow A_{12} $$
Proof: The "N/C" theorem states that $N_G(P)/C_G(P)$ is isomorphic to a subgroup of $\text{Aut}(P)$, so as you have mentioned $$ [N_G(P):C_G(P)] \lvert |\text{Aut}(P)| = \varphi(11) = 10 $$ Hence, $3^2 \mid |C_G(P)|$. In particular, $C_G(P)$ has an element of order 3.
This element necessarily commutes with a generator of $P$, and their product is an element of order 33.
However, $A_{12}$ cannot have an element of order 33 since it would have to be a product of two disjoint cycles of orders 11 and 3 respectively.