Let $G$ be a group of order $240=2^4\cdot 3\cdot 5$.
Assume $G$ is simple, then
- show that there is no subgroup of index 2, 3, 4 or 5
- show that there is no subgroup of index 6
For the first item, the usual method seems to work:
Let $H\leq G$ where $[G:H] = 5$ then let $G$ act on the set of right cosets of $H$ by right multiplication.
Assume $\phi: G\to S_5$ to be its permutation representation. Since $G$ is simple this $\phi$ must be one-to-one which implies $G\cong \mathrm{im}(\phi) \leq G$. But this is impossible since $|S_5| = 5!=120$ while $|G|=240$.
For the second item. Let once again $H\leq G$ such that $[G:H] = 6$. Repeating the same argument does not work here, since $240 = |G| \leq |S_6| = 720$.
Any ideas on how to proceed? I prefer hints to complete solutions.
Perhaps I could use the theorem $[G:H] = [G:K]\cdot [K:H]$ if I would find a $K$ such that $H\leq K\leq G$. (I doubt it though)
Let $\varphi: G \to S_6$ be the permutation representation on the six cosets. This is an embedding since $G$ is simple, so we identify $G$ with a subgroup of $S_6$. Simplicity implies all elements of $G$ must be even permutation. Hence $G\leq A_6$, but $240\nmid 360$.
An off-topic comment: There is in fact no simple group of order $240$.