Let $\phi : G \rightarrow H$ be a group homomorphism, and let $A $, $B$ be subgroups of $G$. If $A$ is normal subgroup of $B$, then prove that $A\phi$ is normal subgroup of $B\phi$
I saw this definiton in my notebook, but there was not any proof for it. Could you give me any hint, theorem or source to prove it?
I know that there must be "work" here. However, I could not think anything. Firstly, I thought that abelian groups can be used but I realized it won't work.
Thanks for every contribution.
To show $A\phi\le B\phi$, use the one-step subgroup test.
Since $A\le G$, we have $e\in A$ and so $e\phi\in A\phi$. Thus $A\phi\neq\varnothing$.
Let $a\phi\in A\phi$, $a\in A$. Then, since $A\unlhd B$, we have $a\in B$, so $a\phi\in B\phi$. Hence $A\phi\subseteq B\phi$.
Suppose $x=r\phi, y=s\phi\in A\phi$, some $r,s\in A$. Then we have
$$\begin{align} xy^{-1}&=(r\phi)(s\phi)^{-1}\\ &=(r\phi)(s^{-1}\phi)\\ &=(rs^{-1})\phi, \end{align}$$
which is in $A\phi$ since $rs^{-1}\in A\le G$.
Hence $A\phi\le B\phi$.
Now suppose $b\phi\in B\phi$ and $a\phi\in A\phi$. Then we have
$$(b\phi)(a\phi)(b\phi)^{-1}=(b\phi)(a\phi)(b^{-1}\phi)=(bab^{-1})\phi,$$
but $bab^{-1}\in A$ as $A\unlhd B$. Hence $(b\phi)(a\phi)(b\phi)^{-1}\in A\phi$.
Hence $A\phi\unlhd B\phi$.