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.



On some endomorphisms of finite groups of odd order
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.


