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?).

  • $\begingroup$ You should take “normal” to mean “equivariant under conjugation”. $\endgroup$
    – k.stm
    Commented Jul 24, 2020 at 16:51

2 Answers 2


Note that any group automorphism has the property that it maps normal subgroups to normal subgroups. In particular, every inner automorphism has this property. On the other hand, what you call a normal endomorphism is exactly an endomorphism which commutes with every inner automorphism. Can you come up with an example of two noncommuting inner automorphisms?

  • 2
    $\begingroup$ This is what I was just about to say. All automorphisms have the property, but are certainly not normal in this sense. $\endgroup$ Commented Jul 24, 2020 at 16:57
  • $\begingroup$ @DavidCraven: Yes, exactly! :) $\endgroup$
    – Mike F
    Commented Jul 24, 2020 at 16:59

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 $a$.


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