# When is the image of a group morphism a normal subgroup?

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. – Mihai Bogdan 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. – Mihai Bogdan Jan 24 '12 at 19:52

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.