I want to solve the following exercise from Dummit & Foote's Abstract Algebra text:
If $A$ is an abelian group with $A \trianglelefteq G$ and $B$ is any subgroup of $G$ prove that $A \cap B \trianglelefteq AB$.
My attempt: Since $A \trianglelefteq G$, it is the kernel of some homomorphism $\varphi:G \to H$. Since $A$ is abelian, any subgroup of it is normal. Thus $A/(A \cap B)$ exists as a quotient group. We define a map $\psi:AB \to H \times A/(A \cap B)$ by $\psi(ab)=(\varphi(b),a(A \cap B))$. It is well defined, for if $ab=a'b'$ then $b'=a'^{-1}ab$ differs from $b$ by an element of $A$, so that $\varphi(b')=\varphi(b)$. Also $ab=a'b'$ gives $a^{-1}a'=bb'^{-1} \in A \cap B$ so that $a(A \cap B)=a'(A \cap B)$. This map is a homomorphism, with kernel $\{ ab \in AB:\varphi(b)=1,a(A \cap B)=A \cap B \}=\{ab \in AB : b \in A,a \in B \}=A \cap B$. Since $A \cap B$ is the kernel of a homomorphism with domain $AB$ we have $A \cap B \trianglelefteq AB$.
Is my solution correct? If not, could you please help me correct it?
Thanks!