# Normal endomorphism on a group

I was learning the Krull-Schmidt theory and came across this concept and just can't understand what's it all about.

A group endomorphism $$f\colon G\to G$$ is called normal iff $$f(aba^{-1})=af(b)a^{-1}$$ for all $$a,b\in G$$. It's true that $$H$$ is a normal subgroup of $$G$$ implies $$f(H)$$ is a normal subgroup of $$G$$, given that $$f$$ is a normal endomorphism on $$G$$.

Is the converse true? E.g. Is it true that "an endomorphism $$f$$ on group $$G$$ images every normal subgroup of $$G$$ to a normal subgroup" implies "$$f$$ is a normal endomorphism"?

If it's not true, some other way to understand this definition would be appreciated(what does it have to do with normality?).

• You should take “normal” to mean “equivariant under conjugation”. Commented Jul 24, 2020 at 16:51

Hint If $$f$$ and $$H$$ are normal, then for all $$g \in G$$ you have $$gf(H)g^{-1}=\{ gf(h)g^{-1} : h \in H \} = \{ f(ghg^{-1}) : h \in H \} \subseteq f(H)$$ Therefore, $$f(H)$$ is normal subgroup.
The converse is not true. The simplest counterexample is $$f: A_5 \to A_5$$ defined by $$f(x)=gxg^{-1}$$ for some $$g\in S_5$$ which we will pick later. Since $$A_5$$ is simple, $$f$$ trivially maps normal subgroups into normal subgroups.
Now, if $$a\in A_5$$ then setting $$b=a$$ you have $$f(aba^{-1})=af(b)a^{-1} \Leftrightarrow \\ gag^{-1}=agag^{-1}a^{-1} \Leftrightarrow \\ (gag^{-1})a=a(gag^{-1})$$
Now, all you have to do is find some $$g\in S_5,a \in A_5$$ such that $$gag^{-1}$$ does not comute with $$a$$. This is easy, pick $$a$$ a 5-cycle, and pick some $$g$$ such that $$gag^{-1}$$ is not a power of $$a$$.