1. ## Quotient groups

Let G be a finite group with $\displaystyle K \lhd G$. If $\displaystyle (|K|,[G : K])=1$, prove that K is the unique subgroup of G having order $\displaystyle |K|$

I was thinking about approaching this problem by..

If $\displaystyle H \leq G$ and $\displaystyle |H|=|K|$, and I figure out what happens to elements of $\displaystyle H$ in $\displaystyle G/K$ then I might be able to see something happen?.... Or maybe I'm just approaching it all wrong

2. Originally Posted by ux0
Let G be a finite group with $\displaystyle K \lhd G$. If $\displaystyle (|K|,[G : K])=1$, prove that K is the unique subgroup of G having order $\displaystyle |K|$

I was thinking about approaching this problem by..

If $\displaystyle H \leq G$ and $\displaystyle |H|=|K|$, and I figure out what happens to elements of $\displaystyle H$ in $\displaystyle G/K$ then I might be able to see something happen?.... Or maybe I'm just approaching it all wrong
suppose $\displaystyle H \leq G$ with $\displaystyle |H|=|K|.$ we have $\displaystyle HK \leq G,$ because $\displaystyle K \lhd G.$ therefore $\displaystyle |G|=n|HK|=n\frac{|K|^2}{|H \cap K|},$ for some integer $\displaystyle n \geq 1.$ so $\displaystyle |H \cap K|[G:K]=n|K|$ and hence $\displaystyle [G:K] \mid n,$ because

$\displaystyle \gcd([G:K],|K|)=1.$ let $\displaystyle n=m[G:K],$ for some integer $\displaystyle m \geq 1.$ then $\displaystyle |K| \geq |H \cap K|=m|K| \geq |K|.$ thus $\displaystyle |H \cap K| = |K|$ and hence $\displaystyle H=K,$ because $\displaystyle |H|=|K|.$