# Prove that there is no subgroup of index $6$ in a simple group of order $240$

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 $240$.
• I don't see why simplicity implies all elements of $G$ to act on $G/H$ as even permutations... Could you elaborate further? Jan 16, 2018 at 7:49
• As example, Let $S = \{1,2,3,4,5,6\}$, why is it impossible for a $g$ to act as a permutation $(1, 2)$? Jan 16, 2018 at 7:59
• You treat $G$ as a subgroup of $S_6$. If $G$ contains an odd permutation, then half of $G$'s element is even, half is odd. Those are even form a subgroup of index $2$, which is a proper normal subgroup of $G$. Jan 16, 2018 at 8:02