# 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.

• 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
• @DerekHolt I think we have $|NH| = |H| |N| / |H \cap N|$ with $H \cap N = N$ because it is easily proved that $N \subset H$. It makes $|NH| = |N|$, and $[NH : N]$ divides $|N|$ and $[G : H]$ which makes it $1$. – freehumorist Jan 6 at 20:55
• @freehumorist As I said in my previous comment, this argument does not work if $G$ is infinite, because in that case $H$ and $HN$ are also infinite. – Derek Holt Jan 6 at 20:59

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.

• @Derek Holt How can you two foresee (or flare) here the equality $[NH : H] = [N : N \cap H]$? (For question no1) – freehumorist Jan 7 at 22:23
• @freehumorist "Forsee or flare"? I'm not sure I understand the verb here, but there's nothing misterious: if we have normality then $\;NH/H\cong N/(N\cap H)\;$ by the second isomorphism theorem (sometimes the third one, depending on the author...) and thus the equality in cardinality (and perhaps you meant this by "flare the equality"). Otherwise we have the bijection $\;f\;$ defined there... – DonAntonio Jan 8 at 8:09