Prove that $AN$ is a subgroup of $G$ if $A$ and $N$ are its subgroups and $N$ is normal in $G$

Question

Let $A$ be a subgroup of a group $G$, and let $N$ be a normal subgroup of $G$ Prove than in this case $AN = \{an|a \in A, n \in N \}$ is a subgroup of $G$

I know that since $N$ is normal, we have $AN = NA$. $1$ clearly lies in $AN$, since both $A$ and $N$ contains $1$. Let $an \in AN$. Then $n^{-1}a^{-1}$ lies in $NA$, hence, it lies in $AN$.

But I'm not sure how to prove that $AN$ is closed under the operation in $G$.

Answer 1

Take $an,bm\in AN$. Since $bm\in AN=NA$, then $bm=m'b'$ for some $m'\in N$ and $b'\in A$. Then $anbm=anm'b'$. Set $r=nm'\in N$. Thus $ar\in AN=NA$, and therefore $ar=r'a'$ for $r'\in N$ and $a'\in A$.

Thus $anbm=anm'b'=arb'=r'a'b'$, where $r'\in N$ and $a'b'\in A$, so $anbm\in NA=AN$ and done.