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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 $?

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

2 Answers 2

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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 $G_1/H_1\cong G_2/H_2$.

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    $\begingroup$ Thank you so much, squared! $\endgroup$
    – Mathtask
    Sep 15, 2022 at 14:31
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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.

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  • $\begingroup$ thank you very much zhang $\endgroup$
    – Mathtask
    Sep 15, 2022 at 14:31

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