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.


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 $H$?

Sure, pick $K=H$.

Specifically, if $N \cong K$, is $\varphi(N) = K$?

Ah, that's more interesting.
If $H$ is finite then indeed this is true, because finite groups don't have proper subgroups isomorphic to them (I assume that $\varphi(N)\subseteq K$ assumption still holds here).
But for infinite groups this does not have to hold. Indeed, consider
$$G=H=\bigoplus_{i=-\infty}^{\infty}\mathbb{Z}$$
$$f:G\to H$$
$$f((x_n))=(x_{n+1})$$
i.e. $f$ is shifting each (bi-directional) sequence to the right. Note that it is an isomorphism. Furthermore take $N=\bigoplus_{i=0}^{\infty}\mathbb{Z}$ which is a normal subgroup of $G$ (well, all are, $G$ is abelian). Then put $K=N$ and note that $f(N)$ (which is equal to $\bigoplus_{i=1}^\infty\mathbb{Z}$) is isomorphic to $K$, and it is a proper subset of $K$.
Note that this counterexamples is in line with "$\varphi(N)\subseteq K$" requirement. Of course it can be modified to show that the opposite proper inclusion may hold as well, simply by taking $N=\bigoplus_{i=-\infty}^0\mathbb{Z}$. Then $f(N)=\bigoplus_{i=-\infty}^1\mathbb{Z}$ which is still isomorphic to $K=N$ but this time $K$ is a proper subset of $f(N)$.
