# problems about normal subgroups and the index

I'm stucked with these problems, could you help me?

I. Let $H$ and $N$ subgroups of a group $G$ such that $N \lhd G$ , $|N| < \infty$, $[G:H] < \infty$ and $([G:H], |N|) = 1$. Prove that $N$ is a normal subgroup of $H$.

Since $N$ is normal in $G$ then $gNg^{-1} = N$ for all $g\in G$ then in particular $hNh^{-1} = N$ for all $h \in H$ so the normality condition is easy, but I don't know how to show that $N < H$.

II. Let $H$ and $K$ be subgroups of a group $G$ such that the index of $H$ in $G$ and the index of $K$ in $G$ are finite. Prove that $[K:H \cap K] = [G:H]$ if and only if $G = HK = KH$.

For the sufficiency I used that $|G|=|HK|=|H||K||H \cap K|$ and $|G/K| = |G|/|K|$. And for the necessity I have no idea but I guess it has something to do with the second isomorphism theorem.

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Please check the wording of the first item. You first define $K$ and then talk about $N$, and I think it should be $([G:N],|H|) = 1$. –  ybungalobill Dec 12 '12 at 9:24
Hint for 1: $|NH:H|$ divides both $|G:H|$ and $|N|$, so must be 1. As for 2, you cannot do calcualtions involving $|G|$, because you are not told that $G$ is finite, and you cannot use the isomorphism theorems, because none of the subgroups involved are known to be normal. So you just have to prove it directly using coset representatives. –  Derek Holt Dec 12 '12 at 9:25
@Amr no, only N is normal in G –  Cybuster Dec 12 '12 at 9:34

1) Already done by Derek holt. Note that $\,[NH:H]=[N:N\cap H]\,$ without assuming any normality of $\,H\,$: just define the following function between the sets of left cosets:

$$f:H^{\backslash NH}\to(N\cap H)^{\backslash N}\,\,,\,\,f(nH):=n(N\cap H)$$

Prove $\,f\,$ is a well-defined bijection (warning: we don't necessarily have group above so $\,f\,$ may not be a homomorphism).

2) (i) If $\, G= HK=KH\,$ , then define similarly as above :

$$f:H^{\backslash G}\to (H\cap K)^{\backslash K}\,\,,\,\,f(gH):=g(H\cap K)$$

$$(a)\;\;\text{Well-defined:}\;\;\;gH=xH\Longrightarrow g^{-1}x\in H . \,\text{Writing}\,\,g=kh\,\,,\,x=k'h'\,\,,\text{we get:}$$

$$h^{-1}k^{-1}k'h'=h_0\in H\Longrightarrow k^{-1}k'=hh_0h'^{-1}\in H\cap K\Longrightarrow k(H\cap K)=k'(H\cap K)\Longrightarrow$$

$$f(gH)=g(H\cap K)=k(H\cap K)=k'(H\cap K)=x(H\cap K)=f(xH)$$

and etc.

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