Simple group problem. I am reading a book of algebra and I have a difficulty in understanding one thing.
In the book it is written there is no simple group of order 528 and the explanation is as follows.
Let $H$ be a Sylow 11-subgroup. Then $n_{11}=12$, which I understood why, and $|N(H)|=44$ and $G$ is isomorphic to a subgroup of $A_{12}$.
Could you please explain?
 A: For the subgroup of $A_{12}$ part, note that $G$ acts transitively by conjugation on the $12$ Sylow $11$-subgroups. Consider the Kernel of this action - it is a normal subgroup and the action is non-trivial, so given $G$ is simple, the Kernel must be the trivial subgroup, and the image is isomorphic to $G$. The image is contained in $S_{12}$ because it consists of permutations of the $12$ subgroups.
Now if the image contained an odd permutation, it would have a subgroup of index $2$ and this subgroup would be normal. But the image is known to be simple, so therefore cannot contain any odd permutations, so must be wholly contained in $A_{12}$.
I have sketched over some parts of this which you should already know. The technique of looking at the action on of $G$ on the Sylow subgroups as a homomorphism to the relevant symmetric group is likely to come up again.
A: There is a group of order 528 - in fact, there are 170 of them up to isomorphism. 
See here:
https://groupprops.subwiki.org/wiki/Groups_of_order_528
