There is no simple group of order 528

Question

I just want to ask whether my argument is right or not? Please if someone can help me.

We assume there is a simple group G of order $528$.

$528=(2^4)(3)(11)$ ,Next we have number of Sylow-$11$ subgroups $n_{11}=12$. ( Since $n_{11}=1$ implies $G$ has a normal Sylow-$11$ subgroup . )

Let $H$ be one of Sylow-$11$ subgroup. Then $|G:N(H)|=12$ ,where $N(H)$ is the normalizer of $H$.

Therefore $|N(H)|=44=2^2(11)$ .

Now $N(H)$ has a proper subgroup, which we call $K$.

$N(H)\subset N(K)$.

This implies $|N(H)|$ divides $|N(K)|$ , $|N(K)|$ divides $528$ ,$|N(K)|>44$ (since $N(H)\subset N(K)$).

Hence $|N(K)|=88$.

Now, $|G:N(K)|=\frac{528}{88}=6$

But $528$ does not divide $6!$

Hence by index theorem $N(K)$ contains a normal subgroup of $G$ .

A contradiction .

Hence $G$ is not simple.

Answer 1

The automorphism group of $H \cong C_{11}$ has order $10$, so there must be an element of order $2$ in $C_G(H)$, and hence $G$ contains an element of order $22$.

The action by $G$ by conjugation on the set of $12$ Sylow $11$-subgroups must be faithful if $G$ is simple, so we have an embedding of $G$ into $S_{12}$.

But $S_{12}$ has no element of order $22$, contradiction. So there is no simple group of order $528$.