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I am very stuck on this proof. I will show the work I have done so far, but it leads me to a dead end... I am not sure how to make the connection between homomorphisms and normality or subgroups and normality.

Show that for homomorphism $\phi:G\to H$, if $G_{1}$ is normal in $G$, then $\phi(G_{1})$ is normal in $H$.

My work: Let $G_{1}\triangleleft G$. Then $gG_{1}=G_{1}g$ for all $g\in G$ or $gG_{1}g^{-1}=G_{1}$. Then there exists a $g_{1}'\in G_{1}$ such that $gg_{1}=g_{1}'g$ or $g_{1}=g^{-1}g_{1}'g$. Since $\phi(G_{1})=\left \{ \phi(g_{1}):g_{1}\in G_{1} \right \}$, then ...

This is where I feel there is a dead end and I am not going anywhere...how do I show that $h\phi(G_{1})=\phi(G_{1})h$ for all $h\in H$?

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    $\begingroup$ This is not true unless $\phi$ is surjective. $\endgroup$
    – J126
    Mar 15 '18 at 19:25
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It is not true, take $G_1=G$ be any non normal subgroup of $H$ , $\phi:G\rightarrow H$ the canonical embedding.

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  • $\begingroup$ The OP assumes that $G_1$ is normal in $G$. $\endgroup$
    – J126
    Mar 15 '18 at 19:23
  • $\begingroup$ @Strants The answer was edited after my comment. $\endgroup$
    – J126
    Mar 15 '18 at 23:16
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$G_1$ is normal in $G$

For all $g\in G$

$gG_1g^{-1} = G_1$

$\phi(g)$ is a homomorphism

If $\phi (g)$ is surjective, then for every $h\in H$ there is some pre-image of $h = \phi(g)$

$\phi(gG_1g^{-1}) = \phi(g)\phi(G_1)\phi(g^{-1}) = \phi(G_1)$

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  • $\begingroup$ Thank you... so $\phi$ must be surjective? It does not say in my class notes... I will email my professor right away! $\endgroup$
    – user482939
    Mar 17 '18 at 20:58

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