Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

This question already has an answer here:

Let $G$ be a group, and $N,H$ be subgroups of $G$ with $N$ normal. Show that $$HN = \{hn \mid h \in H, n \in N \}$$ is a subgroup of $G$.

Thanks in advance!

share|improve this question
    
Check that it satisfies the subgroup axioms. Is $1$ in $HN$? If $a,b \in HN$, is $a b^{-1} \in HN$?. Remember by saying that $a,b \in HN$, you are saying $a = h_1 n_1$ and $b = h_2 n_2$. Let me know if you are still stuck after thinking about this for a while and if so, I'll give a more detailed hint. –  muzzlator Mar 2 '13 at 6:44
add comment

marked as duplicate by Stefan Hansen, Asaf Karagila, Rahul, Seirios, Chris Eagle Mar 2 '13 at 9:42

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

2 Answers

Take $a=hn$ and $b=km$ for $h,k\in H$ and $n,m\in N$ and check whether $ab^{-1}\in HN$. $$(hn)(km)^{-1}=hnm^{-1}k^{-1}=\underbrace{\left(hk^{-1}\right)}_{\in H}\,\,\,\,k\,\,\underbrace{\left(nm^{-1}\right)}_{\in N}\,k^{-1}=\underbrace{\left(hk^{-1}\right)}_{\in H}\,\underbrace{\left(knm^{-1}k^{-1} \right)}_{\in N}$$ Note that the last step holds if and only if $N$ is normal. Thus by the one-step subgroup test, $HN$ is a subgroup if and only if $K$ normalizes $N$.

share|improve this answer
add comment

Hint:

  1. For $H,N\leq G, HN\leq G\iff HN=NH;$

  2. $N\triangleleft G\implies gN=Ng~\forall~g\in G$ and so in particular $\forall~g\in H.$

share|improve this answer
add comment

Not the answer you're looking for? Browse other questions tagged or ask your own question.