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Let $f : G \to G'$ be a group morphism. I need to find a necessary and sufficient condition such that $\operatorname{Im}(f)$ is a normal subgroup of $G'$.

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Hello Mihai! I'm afraid it is not considered polite here to command other users to do something. Your question does not show that you have thought about the problem. Please explain what you've tried so far, and where you are stuck. – Zev Chonoles Jan 24 '12 at 18:35
I'm kind of stuck with {$x\times f(y)/y \in G$} = {$f(y) \times x/y \in G$},$ \forall x \in G'$p.s.: I'm sorry about that. I can't really control the tone of what I write in English. – Bogdan Mihai Jan 24 '12 at 18:44
Mihai, why do you think that there would be anything better than the definition of normality? – Jyrki Lahtonen Jan 24 '12 at 19:15
I think that is the definition of normality. – Bogdan Mihai Jan 24 '12 at 19:52
up vote 4 down vote accepted

To show $\mathrm{Im}(f)$ is normal in $G'$ one can establish that for all $y \in \mathrm{Im}(f)$ and $g \in G'$ one has $gyg^{-1} \in \mathrm{Im}(f)$.

This is equivalent to: for each $x \in G$ and $g \in G'$ there exists some $z\in G$ such that $gf(x)g^{-1} = f(z)$. [This is basically just regurgitating the definition.]

In general, there's not much more that can be said. Given a homomorphism $f:G \to G'$ such that $\mathrm{Im}(f)$ is normal in $G'$, one can always (unless $f$ is the trivial homomorphism) find a bigger group $G''$ containing $G'$ such that $\mathrm{Im}(f)$ is not normal in $G''$. So normality of the image is usually quite sensitive to choice of codomain.

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And of course, im(f) is always normal in its normalizer (or itself), so even if im(f) was not normal in G′, one could shrink G′ to make im(f) normal. – Jack Schmidt Jan 24 '12 at 20:48
Thanks Jack! I forgot to mention the other direction. :) – Bill Cook Jan 24 '12 at 20:49
In other words, to guarantee that Im(f) is a normal subgroup of G′, G' must be Im(f), right? Thanks! – Bogdan Mihai Jan 24 '12 at 22:16

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