$A_6$ does not have a subgroup of order $120$

Question

I was reading the following proof of why there is no simple group of order $120$:

A group of order $120$ cannot be simple

And I couldn't understand the following: "so $A_6$ has a subgroup of order $120$ which is impossible".

Why is it impossible? Isn't the order of $A_6$ $6! / 2$, which equals $360$?

And since $120$ divides $360$, I can't see what's the problem in it...

Answer 1

I will recite the theorem given in the linked question:

Theorem: If a simple group $G$ has a proper subgroup $H$ such that $[G:H]=n$ then $G\hookrightarrow A_n$.

Now, Assume you have a subgroup $H \le A_6$, $|H|=120$. The group $A_6$ is simple and $[G:H] = 3$. Thus, by the theorem above, there is an embedding $A_6\hookrightarrow A_3$, a contradiction.

Answer 2

I am unable to find a proper reference for the theorem I am referring to, so here's the relevant part from page 4 of Isaacs' "Finite Group Theory" book.