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?