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??
cany you help in explaining these three things?