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Let $G$ be a group of odd order. It is known that if every central automophism of $G$ acts trivially on the center, then $G$ is purely non-abelain, this amounts to saying that every central endomorphism $u$ of $G$ (an endomorphism such that $x^{-1}u(x) \in Z(G)$, for all $x \in G$) is an automorphism.

I could generalize this (using some ring theory) to the case of the automorphisms acting trivially on a quotient of an abelain normal subgroup: Let $A$ be an abelian normal subgroup of $G$. If every automorphism of $G$ acting trivially on $G/A$ leaves $A$ elementwise fixed, then every endomorphism of $G$ acting trivially on $G/A$ is an automorphism.

I wonder if one can see a straightforward purely group theoretic proof of this result.

Thanks in advance.

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    $\begingroup$ Surely, if $N$ is nonabelian, then the inner automorphism of $G$ induced by conjugation by a non-central element of $N$ will act trivially on $G/N$ but not on $N$, so your hypotheses cannot hold. $\endgroup$ – Derek Holt Oct 3 '13 at 14:11
  • $\begingroup$ You are right. I will remove the last question, to concentrate on the first one. $\endgroup$ – Yassine Guerboussa Oct 3 '13 at 14:52
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You are assuming that $A$ is an abelian normal subgroup of a group $G$ of odd order, with the property that all automorphisms $\alpha$ of $G$ that induce the identity on $G/A$ act trivially on $A$. You want to prove that all endomorphisms of $G$ that induce the identity on $A$ are automorphisms.

I will prove the contrapositive of that statement. Suppose that there exists an endomorphisms $\phi$ of $G$ with nontrivial kernel $K$, where $\phi(A)=A$ and $\phi$ induces the identity on $G/A$. I will prove your assertion by constructing an automorphism of $G$ that induces the identity on $G/A$ but not on $A$.

Clearly $K \le A$. Let $H = {\rm im}(\phi)$. If $H \cap K \ne 1$, then $\phi^2$ still induces the identity on $G/A$ and has smaller image than $\phi$. So, by replacing $\phi$ by a power, we may assume that $H \cap K = 1$ and hence $H$ is a complement to $K$ in $G$.

Now we can define an automorphism of $G$ that fixes $H$ and $K$, induces the identity on $H$ and inverts every element of $K$. Since $K \le A$, this automorphism acts trivially on $G/A$, but not on $A$, so this proves your statemen.

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  • $\begingroup$ You said $\phi$ is an endomorphism that acts trivially on $G/A$ and $A$ and has a non trivial kernel. This can not hold, as $\phi(x)=1$ implies $x^{-1} \phi(x) = x^{-1} \in A$, thus $x \in A$ and so $\phi(x)=x=1$. $\endgroup$ – Yassine Guerboussa Oct 3 '13 at 16:53
  • $\begingroup$ Yes, so $x \in A$ and hence $K \le A$, which is what I wrote. Why should that imply $x=1$? I didn't say that $\phi$ acts trivially on $A$. I am proving the contrapositive of what you asked. I am assuming that there is an endomorphism of $G$ that acts trivially on $G/A$ that is not an automorphism (which is certainly possible in general!) and deducing that there must be an automorphism of $G$ that induces the identity on $G/A$ but not on $A$. $\endgroup$ – Derek Holt Oct 3 '13 at 20:08
  • $\begingroup$ Yes I see. I'm confused when you are saying $\phi$ fixes $A$, you just meant that $A$ is $\phi$-invariant. Note that this is included in our statement that $\phi$ acts trivially on $G/A$. $\endgroup$ – Yassine Guerboussa Oct 3 '13 at 20:40
  • $\begingroup$ This doesn't seem right at all. $\phi(A)=A$ and $\ker(\phi)\leq A$ is impossible for $A$ finite. And there is no guarantee that the image of a higher power of $\phi$ and $K$ will generate $G$. And in your last sentence you can't fix $K$ and invert every element of it at the same time. Unless it has exponent 2, where those are the same. $\endgroup$ – zibadawa timmy Jan 11 '14 at 12:13

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