Can someone please check my solution to the following question please.
Let $A$ and $B$ $N$ be normal subgroups of a group $G$ such that $N\subset A$, $N \subset B.$ If $G=AB$ and $A\cap B=N,$ prove that $G/N\cong A/N\times B/N\\$
$\textbf{Proof:}$ Let the function $f:G/N\rightarrow A/N \times B/N$ be defined by $f(abN)=(aN, bN)$. To show $f$ is an isomorphic map, we have to show it is well-defined homomorphic bijective map.
Well-definedness
Let $abN=a'b'N,$ which is equivalent to $ab\equiv a'b' \pmod N$, then $a'^{-1}ab\equiv b' \pmod N,$ So $a'^{-1}a\equiv 1\pmod N$ and $b\equiv b' \pmod N$ and so $a'^{-1}a\in N$ which gives $a\equiv a' \pmod N$ and $b\equiv b' \pmod N$. Hence $aN=a'N$ and $bN=b'N$. Therefore $(aN, bN)=(a'N, b'N).$ This shows $f$ is well-defined.
Homomorphism:
Let $a, a', b, b' \in G$, then $f(aa'bb'N)=(aa'N, bb'N)=(aNa'N, bNb'N)=(aN, bN)(a'N, b'N)=f(abN)f(a'b'N)$ This shows that $f$ is a homomorphic mapping.
Injectivity
Suppose $f(abN)=f(a'b'N)$, then $(aN, bN)=(a'N, b'N)$. Since $(a'N, b'N)(a'N, b'N)^{-1}=(a'N, b'N)(a'^{-1}N, b'^{-1}N)=(N,N)$, then $(aa'^{-1}N, bb'^{-1}N)=(N,N).$ But $aa'^{-1}N=N$ and $bb'^{-1}N=N$ means both $aa'^{-1}, bb'^{-1}$ are in $N$, and so $aN= a'N$, $bN=b'N$. So $aNbN= a'Nb'N=a'b'N= abN\\$ Hence $f$ is injective.
Surjectivity
Suppose $(aN, bN)\in A/N \times B/N$ for any $a\in A, b\in B$, since $G=AB$, then for any $g\in G$ where $g=ab\in AB$. Then $a=gb^{-1},$ $b=a^{-1}g$. So if $(aN, bN)\in A/N \times B/N$ implies there exist $gb^{-1}\in A$ and $a^{-1}g\in B$ so that $gb^{-1}a^{-1}gN\in G/N$ for which $f(abN)=(aN,bN)$. So $f$ is surjective.
Thank you in advance.