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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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    I think that is the definition of normality. – 2012-01-24

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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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    In other words, to guarantee that Im(f) is a normal subgroup of G′, G' must be Im(f), right? Thanks! – 2012-01-24