Let $\phi:G_1\to G_2$ isomorphism of groups. Let $H_1\lhd G_1,H_2\lhd G_2$ and suppose that $\phi (H_1) = H_2$. Is true that $G_1/H_1 \cong G_2/H_2$?

Question

Let $\phi : G_1 \rightarrow G_2$ isomorphism of groups. Let $H_1 \lhd G_1, \; H_2 \lhd G_2$ and suppose that $\phi (H_1) = H_2$. Is true that $G_1/H_1 \cong G_2/H_2$?

What I think, is that $\ker(\phi)$ can be greater than $H_1$.

Can I create a surjective homomorphism $\alpha: G_1 \rightarrow G_2/H_2$? I can also say that $\ker(\alpha) = H_2 = \phi $?

Answer 1

Let $q_2:G_2\to G_2/H_2$ be the quotient map $g\mapsto gH_2$.

The composition $q_2\circ\phi:G_1\to G_2/H_2$ is a surjective group homomorphism. Observe — utilizing the assumption $\phi[H_1]=H_2$ — that $$x\in\ker(q_2\circ\phi)\iff \phi(x)\in H_2\iff x\in H_1$$ i.e., $\ker(q_2\circ \phi)=H_1$. Finally, apply the first isomorphism theorem to conclude $G_1/H_1\cong G_2/H_2$.

Answer 2

Let $p_i$ be the projection from $G_i$ to $G_i/H_i$ for $i=1,2$. Since for every $g,g'$ in $G_1$, $p_1\left(g\right)=p_1\left(g'\right)$ implies that $gH_1=g'H_1$ and hence $\phi\left(g\right)H_2=\phi\left(g'\right)H_2$, which implies that $p_2\circ\phi \left(g\right)=p_2\circ\phi \left(g'\right)$, there shall be a unique induced mapping $\phi'$ from $G_1/H_1$ to $G_2/H_2$. $\phi'$ is surjective since $p_2\circ\phi$ is surjective. Suppose that $gH_1,g'H_1\in G_1/H_1$, then if $\phi' \left(gH_1\right)=\phi' \left(g'H_1\right)$, then $p_2\circ\phi\left(g\right)=p_2\circ\phi\left(g'\right)$ and hence $\phi\left(g\right)H_2=\phi\left(g'\right)H_2$, take $\phi^{-1}$ on both sides and hence $gH_1=g'H_1$, which implies that $\phi'$ is bijective.