Isomorphism between groups and subgroups.

Question

Say I know that $G_1,G_2$ are isomorphic groups, both with normal subgroups $N_1,N_2$ respectively. Can I take an isomorphism $\psi$ between $G_1,G_2$ s.t $\psi(N_1)=N_2$? Meaning that $\psi$ also defines a isomorphism between $N_1$ and $N_2$?

The thm I'm trying to prove is if $G_1 \cong G2, N1\cong N2,N1\unlhd G1, N2 \unlhd G2$ then $G1/N1 \cong G2/N2$, and proving the above statement I would be able to use the first isomorphism theorem and that's it. Is it even true?

Any hint would be helpful on how to approach this problem.

Answer 1

Let $\psi\colon G_1\rightarrow G_2$ be a group isomorphism and $N_1$ be a normal subgroup of $G_1$. Then of course $N_2=\psi(N_1)$ is a normal subgroup of $G_2$. However, the normal subgroups of a group may have different cardinality in general, so that not always $\psi(N_1)=N_2$ for arbitrary normal subgroups $N_1$ and $N_2$ in $G_1$ respectively $G_2$.

Answer 2

Well, no. For example, take $G_1=G_2=N_1=S_3$ and $N_2=\langle (123)\rangle$. No isomorphism will satisfy $\psi(N_1)=N_2$, because they don't have the same cardinality.

Anyway, the statement you want to prove is false. If we take $G_1=G_2=\mathbb{Z}$ and $N_1=2\mathbb{Z}, N_2=3\mathbb{Z}$ then $G_1\cong G_2$ and $N_1\cong N_2$. However, $G_1/N_1$ and $G_2/N_2$ can't be isomorphic, as they have different cardinalities.

Answer 3

It is not true, consider $G_1=G_2=C_4\times C_2$. Let $N_1=0\times C_2$ and $N_2 =H\times 0$ where $H$ is the subgroup of order $2$ in $C_4$. Then $N_1\cong N_2$, but $C_4=G_1/N_1\not\cong G_2/N_2=C_2\times C_2$