# If $G$ is a simple group with order greater than $60$, then $G$ has no proper subgroup of index less-than-or-equal-to $5$.

I want to understand a proof for: If $G$ is a simple group with order greater than $60$, then $G$ has no proper subgroup of index less-than-or-equal-to $5$.

If we let $G$ be a simple group with order greater than $60$ and let $H$ be a subgroup of $G$, we can note that the index of $H$ in $G$ cannot be $1$ or else it would contradict that $G$ is simple (it contradicts that it is proper).

Then, we could let the index be $2$, but if $2$ is the smallest prime dividing $|G|$ then $H$ is normal in $G$. (First of all, how do we know this?) Then this also contradicts the simplicity of $G$.

Then, we could let the index be $3$. But if $3$ is the smallest prime dividing the order of $G$, we have $H$ being normal again which contradicts simplicity. Thus, we may assume that $2$ divides the order of $G$.

...I don't quite understand this proof so far. Is it saying that any prime cannot equal the order of $G$? (If any prime did, then wouldn't it be the smallest and lead to the contradiction of simplicity? Or was that only for 2 and 3? Is this is a general rule that a simple group of prime order has a normal subgroup?)

And, certainly some odd numbers are not prime, so how do we know the order of $|G|$ cannot be odd?

...there is more to this proof, but I just want to understand first things first :) it will go on to check why the index of $H$ in $G$ cannot be $4$ or $5$, so it's a brute-force problem. I just understand the bit about the smallest prime dividing $G$ implying that $H$ is normal?

If $G$ has a subgroup of index $k$, then the action of $G$ on the cosets of $K$ induces a homomorphism from $G$ to $S_k$ whose image is a transitive subgroup $H$ of $S_k$. If in addition, $G$ is simple then $G$ must be isomorphic to $H$. It's easy to see that if $|G|>2$ then $H\le A_k$.
• Lord Shark the Unknown, your answer is really helpful to me. I understand that concept of mapping to $S_{n}$ which I struggled with for awhile, so that's wonderful. But I still don't understand what the answer is to the question about the primes or odds here - can $G$ have prime order, and what does the "odd" part have to do with anything (since $2$ must divide the order of $G$, I guess the order can't be odd?) – PBJ Dec 13 '17 at 18:41
• Oh...because $G$ is a group of $A_{k}$? I think I partially understand this part, but not fully. – PBJ Dec 13 '17 at 18:42