Mathematics Asked by DarkGlimmer on January 1, 2022

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 $fcolon Gto G$ is called * normal* iff $f(aba^{-1})=af(b)a^{-1}$ for all $a,bin G$. 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?).

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?

Answered by Mike F on January 1, 2022

**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 $gin S_5$ which we will pick later. Since $A_5$ is simple, $f$ trivially maps normal subgroups into normal subgroups.

Now, if $ain A_5$ 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 $gin S_5,a in A_5$ 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 $a$.

Answered by N. S. on January 1, 2022

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