Group of order $1320$ is not simple

Question

Group of order $1320 = 2^3\cdot 3\cdot 5\cdot 11$ is not simple.

Proof. Suppose there is a simple group $G$ of order $1320$. Then the number of Sylow $11$ subgroup $n_{11} = 12$. Let $G$ act on ${\rm Syl}_{11}(G)$ by conjugation so we get an injective homomorphism $G\to S_{12}$. Now, if $P_{11}\in{\rm Syl}_{11}(G)$ then $|N_G(P_{11})| = 110$ and $|N_{S_{12}}(P_{11})| = 110$.

I want to show the map $\varphi:G\to S_{12}$ equals $G\xrightarrow{\tilde{\varphi}} A_{12}\hookrightarrow S_{12}$ for some $\tilde{\varphi}$ so that I can get a contradiction $|N_{A_{12}}(P_{11})|= 55$. How can I show this?

Answer 1

Note that the image of $\varphi$ is contained in $A_{12}$. Indeed, if it wasn't then we would have a surjective map $G\to S_{12}\to\{1,-1\}$, by taking the composition of $\varphi$ with the sign homomorphism. But clearly there can't be a surjective homomorphism $G\to\{1,-1\}$, because $G$ is simple.

Answer 2

Here is a proof from $1900$: A simple group of order $1320 = 2^3\cdot 3\cdot 5\cdot 11$ would have $12$ conjugate subgroups of order $11$ and could be represented as a doubly-transitive group of degree $12$. However, there is only one doubly-transitive group of this order and degree, and it is "composite" (see the paper by G. A. Miller 1899).

Reference: Proof that there is no Simple Group whose Order Lies Between 1092 and 2001.