If $\varphi: G \to H$ is an isomorphism and $N \trianglelefteq G$, then, $\varphi(N) \subseteq K$, where $K$ is some normal subgroup in $H$? If $\varphi: G \to H$ is an isomorphism and $N \trianglelefteq G$, then, is it true that $\varphi(N) \subseteq K$, where $K$ is some normal subgroup in $H$? Specifically, if $N \cong K$, is $\varphi(N) = K$?
Motivation:
Suppose that $\pi: G \to H/K$ such that $\pi (g) = \varphi (g) K$, where $\varphi$ is some isomorphism from $G$ to $H$. Also, that $\text{ker}(\pi) \subseteq N$. If $\alpha: G/N \to H/K$ such that $\alpha (gN) = \pi (g)$. My question: Is $\alpha$ well-defined? For more details click on this post.
My attempt:
If $a, b \in G$ and $aN = bN \Rightarrow (a * b^{-1})N = N \Rightarrow a * b^{-1} \in N \Rightarrow \varphi (a * b^{-1}) \in K$ (here's when I get stuck and I all I wanted to prove was that $\varphi (a * b^{-1}) \in K$).
Hope my question is clear. Thanks.

In order to get $\varphi(N)$ normal in $H$ (so that you can take $\varphi(N)$ as the sought $K$), your $\varphi$ is even overdetermined. It suffices to assume $\varphi$ a surjective homomorphism. In fact in this case, $\forall h\in H$, $\exists g\in G$ such that:
\begin{alignat}{2}
h\varphi(N)h^{-1} &= &&\{h\varphi(n)h^{-1}, n\in N\} \\
&= &&\{\varphi(g)\varphi(n)\varphi(g)^{-1}, n\in N\} \\
&= &&\{\varphi(g)\varphi(n)\varphi(g^{-1}), n\in N\} \\
&= &&\{\varphi(gng^{-1}), n\in N\} \\
&\stackrel{(N\unlhd G)}{=} &&\{\varphi(n'), n'\in N\} \\
&= &&\varphi(N)
\end{alignat}
whence $\varphi(N)\unlhd H$. Of course this holds, a fortiori, if $\varphi$ is an isomorphism (just replace "$\exists$" with the stronger "$\exists !$").
