On the 11 Sylow subgroup of a group of order 792 It seems that the 11-Sylow subgroup of a group of order $792=11×2^3×3^2$ is normal. Could you help me why it is true
If there are $12$ Sylow $11$-subgroups, then the normalizer of a Sylow $11$-subgroup has order $66$. Since $3$ does not divide $11-1=10$, an element of order $3$ centralizes an element of order $11$, giving an element of order $33$, but there is no such element in $S_{12}$.

Sorry, for some reason I thought we were just trying to prove that there was no simple group of order $792$.

OK, carrying on, the element of order $3$ must be in the kernel of the map $\phi:G \to S_{12}$. Now $\phi$ is the action by conjugation on the set of $12$ Sylow $11$-subgroups, which is transitive. Also a Sylow $11$-subgroup does not normalize any other Sylow $11$-subgroup, so its image consists of an $11$-cycle. So the image of $\phi$ is a doubly transitive group. Since $3$ divides $|\ker(\phi)|$, we have $|{\rm im}(\phi)|=132$ or $264$.

If $|{\rm im}(\phi)|=264$, then the element of order $2$ that normalizes the Sylow $11$-subgroup has $5$ $2$-cycles, so it is an odd permutation. So intersecting with $A_{12}$ gives us a group of order $132$.

Either way, we have a group of order $132$ with $12$ Sylow $11$-subgroups. It is actually a $2$-transitive Frobenius group, which is impossible, because it does not have prime power degree. But arguing directly, the counting argument suggested by Ronno starts to kick in. We have $12$ elements of order $11$, leaving only $12$ other elements. So either there must be a unique (and hence normal) Sylow $2$-subgroup or a unique Sylow $3$-subgroup. Either way, it would be centralized by the Sylow $11$-subgroup, and this time we really do get a contradiction!!!
Breaking 792 into prime divisors, we get $792$ = $11*3^2*2^3$.
That means we have 3 Sylow p-groups of order $11$, $3^2$ and $2^3$ by the first Sylow theorem.

Let $n_{11}$ represent the Sylow p-group of order $11$

Let $n_2$ represent the Sylow p-group of order $2^3$

Let $n_3$ represent the Sylow p-group of order $3^2$

Using $n_{11}$ would be the easiest because it's the only order that is prime. So by the third Sylow theorem, $n_{11}$ $\equiv 1$ mod $11$. We know that $n_{11}$ | $11*3^2*2^3$

This would imply that $n_{11}$ = $1$. Thus the Sylow p-subgroup of order $11$ is a normal subgroup.