Quotient groups

**ux0** · October 28th 2009, 05:48 PM

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

I was thinking about approaching this problem by..

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

**NonCommAlg** · October 28th 2009, 06:49 PM

Originally Posted by ux0

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

I was thinking about approaching this problem by..

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

suppose $H \leq G$ with $|H|=|K|.$ we have $HK \leq G,$ because $K \lhd G.$ therefore $|G|=n|HK|=n\frac{|K|^2}{|H \cap K|},$ for some integer $n \geq 1.$ so $|H \cap K|[G:K]=n|K|$ and hence $[G:K] \mid n,$ because

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

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