What it says on the tin.

A group endomorphism $v\colon G\to G$ is called normal if $v(aba^{-1})=av(b)a^{-1}$ for all $a,b\in G$. Equivalently, the map $g\mapsto v(g^{-1})g$ is a group homomorphism. Equivalently, the image of this map commutes with the image of $v$.

$\operatorname{End}(G)$, all endomorphisms of G, is a monoid under composition. Let $M$ be the set of all normal endomorphisms of $G$. This forms a submonoid. It is normal if $vM=Mv$ for all $v\in\operatorname{End}(G)$. So is it normal?

If we restrict to automorphisms, the normal automorphisms form what are also known as the central automorphism group. This is known to be a normal subgroup, and to my understanding the proof relies on the center being characteristic. This doesn't generalize to endomorphisms, though.

  • $\begingroup$ Let $f_a$ denotes the conjugation by $a$ in $G$, notice that $f_a \in End(G)$ and set of all $f_a$ constitutes $Inn(G)$ which is subgroup of $End(G)$. we can say that if $v:G\to G$ is normal if $f_a\circ v=v\circ f_a$ so $M=C_{End(G)}Inn(G)$ and $M$ is normal in $N_{End(G)}{Inn(G)}$. Thus, if $M$ is normal in $End(G)$ then $Inn(G)$ is normal in $End(G)$ so your question is equivalent to say that "Does $Inn(G)$ have to be normal in $End(G)$ ? " Am I right ? $\endgroup$ – mesel Apr 12 '14 at 7:26
  • $\begingroup$ @mesel I don't get the "thus" part. Before then, yes, but I don't think the rest is true. $\endgroup$ – zibadawa timmy Apr 12 '14 at 21:23

No, it's not.

First the abstract reason, followed by an example.

First, $v\colon G\to G$ is normal if and only if $Sv*id$ is a normal endomorphism, where $S$ is the inversion map. This is in turn equivalent to the images of $v$ and $Sv*id$ commuting.

Suppose $G$ is indecomposable. Then $id=v*(Sv*id)$ implies that one of the two morphisms is a central automorphism. Since their images commute, the other is thus central.

Let $w\colon G\to Z(G)$. This gives a normal endomorphism. Then for any endomorphism $v$ the composition $wv$ is also a normal endomorphism with central image. So we want to know if there is a normal endomorphism $w'$ (depending on $v,w$) such that $vw'=wv$. If we can find $v,w$ with no such $w'$, then we have the claim

The obstruction is intuitively clear, though slightly bothersome to write out.

More specifically, if we had a $v$ such that $v(Z(G))\setminus Z(G) = v(Z(G))\setminus 1$; and a non-trivial central homomorphism $w$ with $wv$ nontrivial, then no such $w'$ can exist: necessarily $vw'$ is not central.

Using GAP I found the following example, though it may not be of minimal order.

$G=\textrm{SmallGroup}(32,11)=(C_4\times C_4)\rtimes C_2$. As a permutation group $G$ can be generated by $x=(1,5)(2,6)(3,4)(7,8)$ and $y=(1,8,7,3,4,2,6,5)$. The center is generated by $(1,6,4,7)(2,3,8,5)$.

Define $w$ by $w(x)=1$, $w(y)= (1,6,4,7)(2,3,8,5)$.

Define $v$ by $v(x)=1$, $v(y)=(2,3,8,5)$.

Then $w,v$ have the desired properties, and we conclude that the normal endomorphisms of $G$ are not a normal sub-monoid.


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