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If $N$ is a normal subgroup of $G$ and $M$ is a normal subgroup of $G$, and if $MN=\{mn|m\in M,n\in N\}$, prove that $MN$ is a subgroup of $G$ and that $MN$ is a normal subgroup of $G$.

The attempt: I tried just starting by showing that $MN$ is a subgroup of $G$. I said let $a=m_1 n_1$ for some $m_1 \in M$ and $n_1 \in N $ and let $b=m_2 n_2$ for some $m_2 \in M$ and $n_2 \in N$, and we need to show $a*b^{-1}$ $\in MN$.

So I get $a*b^{-1}$=$m_1n_1n_2^{-1}m_2^{-1}=m_1n_3m_2^{-1}$ but then I don't know how to show that this is in $MN$. Tips on this or the next part of the problem?

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    Now, you will need to use your hypothesis, $N$ is a normal subgroup. Recall $N$ normal in $G$ means for each $x\in G$, we have $xN=Nx$. So we have $m_1( n_3 m_2^{-1})= m_1( m_3 n_3)=m_3 n_3$2012-11-12

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From the last line you wrote: $a*b^{-1}=m_1n_3m_2^{-1}=m_1m_2^{-1}m_2n_3m_2^{-1}$

Do you see why this is $MN$?

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    @AllisonCameron Good work :)2012-11-12
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$m_1n_3m_2^{-1}=m_1m_2^{-1}(m_2n_3m_2^{-1})\in MN$

General Lemma: if $\,M,N\,$ are subgroups of $\,G\,$ , $\,MN\,$ is a subgroup iff $\,MN=NM\,$ .

In particular, if $\,M\triangleleft G\,$ or$\,N\triangleleft G\,$ ,then $\,MN=NM\,$

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    Thank you! Can't believe I didn't see that!2012-11-12