Defining set-theoretic section on finite abelian group

Question

Let $A$ be a finite abelian group and $\cdot 2:A \to 2A$ the square-epimorphism. Let $\{ a_i\}$ be a minimal generating set for $A$, s.t. $2a_i \neq 0$. I want to define a set-theoretic section $s:2A \to A$, s.t. $s(2a_i)=a_i$. Is there a possibility to extend this multiplicatively somehow, so that $s (\sum_i r_i 2 a_i) = \sum_i r_i a_i $ for $0 \leq r_i < |2a_i|$, by choosing particular words on letters $a_i$ for every element $2a \in 2A$?

Answer 1

$\newcommand{\Set}[1]{\left\{ #1 \right\}}$I was writing up a comment, but since it was a bit too long, I have expanded it into an answer.

The kernel of $\cdot 2$ would be $K = \Set{ a \in A : 2 a = 0 }$, and then $\cdot 2$ yields an isomorphism $A / K \to 2 A$. Assuming none of the $a_{i}$ is in $K$, $\cdot 2$ restricts to a bijection on the set $$ B = \Set{ \sum_{i} r_{i} a_{i} : 0 \le r_{i} < N_{i}/2}, $$ where $N_{i}$ is the order of $a_{i}$, so that $N_{i}/2$ is the order of $2 a_{i}$. This is simply because if $$ \sum_{i} 2 r_{i} a_{i} = \sum_{i} 2 s_{i} a_{i}, $$ with $0 \le r_{i} , s_{i} < N_{i}$, we have $2 r_{i} = 2 s_{i}$ and thus $r_{i} = s_{i}$.

So, yes, you can construct your $s$ this way, as the inverse of this bijection.

This works in general to get a map $s$ such that composing $s$ and then $\cdot 2$ you have the identity on $2 A$, only it will not have in general the property $s(2 a_{i}) = a_{i}$ you asked for, in case $a_{i}$ is an involution.