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?