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Let $A$ be an abelian group and $B$ a subgroup of $A$. Suppose $\sigma: A \to A$ is such that $\sigma(B) \subseteq B$, then it induces a map $\tilde{\sigma}: A/B \to A/B$.

I am interested in finding simple counterexamples (or proving the corresponding results, although I believe counterexamples should exist) to the following questions:

  1. Given $\tau \in \text{End}(B)$, does there exist $\sigma \in \text{End}(A)$ such that $\sigma|_B = \tau$?

  2. Given $\tau \in \text{End}(A/B)$, does there exist $\sigma \in \text{End}(A)$ such that $\tilde{\sigma} = \tau$?

  3. Given $\tau \in \text{Aut}(B)$, does there exist $\sigma \in \text{Aut}(A)$ such that $\sigma|_B = \tau$?

  4. Given $\tau \in \text{Aut}(A/B)$, does there exist $\sigma \in \text{Aut}(A)$ such that $\tilde{\sigma} = \tau$?

A counterexample for Question 3 has already been constructed in this question, which is quite clever. It was after seeing this counterexample that I was prompted to ask about the other cases, since a simple search did not lead me to find counterexamples for the other three questions on MSE. I think this is a very natural question and should be asked by someone.

Specifically, I am not sure if this problem can be viewed from a "higher perspective" using homological algebra methods.

Zhang Yuhan
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    Please focus on one question only, see here. The linked dupe (that I found with Approach Zero) answers questions 1 and 2, and you already answered 4. Maybe the given answer to 2 also answers 3, but I didn't check. In any case it should be separate. But probably there is also an exact duplicate of 3. – Martin Brandenburg Nov 06 '24 at 07:30
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    Homological interpretation: only possible if $B$ is fully invariant so that your question is about a surjectivity of certain maps. But fully invariant subgroups of abelian groups are rare. Also the maps are not additive. So it's unlikely. – Martin Brandenburg Nov 06 '24 at 07:39
  • True. The linked examples solve all questions. One only need to figure out all endomorphism of $\mathbb Z_4 \times \mathbb Z_2$ and $\mathbb Z_2 \times \mathbb Z_2$, which is easy. – Zhang Yuhan Nov 06 '24 at 08:03

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