# 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 $G_1/H_1 \cong G_2/H_2$?

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 $$\ker(\alpha) = H_2 = \phi$$?

• Please note that the supposed counterexamples here are for surjective $\varphi$, not necessarily an isomorphism. Sep 15, 2022 at 13:46
• But if it's an isomorphism it might change something in the kernel of $\phi$? Sep 15, 2022 at 13:48
• Since $\phi$ is an isomorphism, $\ker(\phi)$ certainly isn't very big... Definitely not greater than $H_1$, even if $H_1$ were trivial. Sep 15, 2022 at 13:52
• Try the map $xH_1 \mapsto \phi(x)H_2$. Sep 15, 2022 at 13:55
• Can I create a surjective homomorphism $\alpha: G_1 \rightarrow G_2/H_2$? I can also say that $\ker(\alpha) = H_2 = \phi$? Sep 15, 2022 at 14:10

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 $$G_1/H_1\cong G_2/H_2$$.
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.