If $\varphi: G \to H$ is an isomorphism and $N \trianglelefteq G$, then, $\varphi(N) \subseteq K$, where $K$ is some normal subgroup in $H$?

Question

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.

Answer 1

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 "$\exists !$").

Answer 2

I think proving $\varphi N\trianglelefteq H$ be enough to answer your question.

If $h\in H, x\in N$, then I'll let $g=\varphi^{-1}h$. Then $h\cdot\varphi x\cdot h^{-1}=\varphi g\cdot\varphi x\cdot(\varphi g)^{-1}=\varphi g\cdot\varphi x\cdot\varphi (g^{-1})=\varphi(gxg^{-1})$ holds. Since $gxg^{-1}\in N$, so $hxh^{-1}\in\varphi N$.

It holds for arbitaty $x\in N, h\in H$, so $\varphi N\trianglelefteq H$.

Answer 3

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 $f(N)$.