TITLE: prove : if $G$ is a finite group of order $n$ and $p$ is the smallest prime dividing $|G|$ then any subgroup of index $p$ is normal
QUESTION [10 upvotes]: Prove:
If $G$ is a finite group of order $n$ and $p$ is the smallest prime dividing $ |G| $ then any subgroup of index $p$ is normal
where  $ |G| $ is the order of $G$

This is a result in Abstract Algebra by Dummit and Foote at page 120 .
The proof is produced but there is some points which is not obivous for me !
First, in page 121 in the proof, it says, all divisors of $ (p-1)!$ are less than $p$.
why this is true ? can any one explain?
Second, why does "every prime divisor of $ k$  is greater than or equal to $p$ "force that k=1 ??
Why does $ k = 1 $ under this condition??
Third, if $ k=1 $
Then, the order of $  K$  = the order of $H$
Why does this mean that $ K=H $  in this case??
Can you help in explaining these three things?

REPLY [4 votes]: We can use the Cayley Theorem to show that:

Theorem: Let the group $G$ have a subgroup $H$ of index $n$, then there is a normal subgroup $K$ of $G$ such that $$K\subseteq H,~~ m=[G:K]<\infty,~~m\mid n!$$

Now let $H\leq G$ and $[G:H]=p$ such that $p$ has the property the question noted. So according to the above theorem there is a normal subgroup $K$ of $G$ contained in $H$ such that $[G:K]\big|p!$. But $[G:H]\big ||G|$ also, so since $p$ is the smallest dividing number then we have necessarily have $$[G:K]\big|p$$ and this means that $K=H$.