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I didn't see this problem before. I motivated by the questions

  1. Is every commutative group structure underlying at least one (unitary, commutative) ring structure

  2. A basic question about rings

Suppose A is an Abelian group such that it is possible to define an associative unitary ring structure on A. An element $x\in A$ is called potentially identity if there exists an associative ring $R$ such that $A$ is the additive group of $R$ and $x=1_R$, the identity element of $R$. Determine the set of all potentially identity elements of $A$.

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  • $\begingroup$ Note that if a potential identity $x$ has finite order $n,$ then the group $A$ must have exponent $n.$ $\endgroup$
    – Geoff Robinson
    Commented Jan 20, 2014 at 20:27
  • $\begingroup$ @GeoffRobinson: Very good point! $\endgroup$
    – Sh.M1972
    Commented Jan 20, 2014 at 20:30
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    $\begingroup$ @GeoffRobinson: Actually, that gives a complete answer when $x$ has finite order $n$. If $A$ has exponent $n$, then it's a $\mathbb{Z}/n\mathbb{Z}$-module, and the subgroup $\langle x\rangle\cong\mathbb{Z}/n\mathbb{Z}$ generated by $x$ is injective as a $\mathbb{Z}/n\mathbb{Z}$-module, so $A=\mathbb{Z}/n\mathbb{Z}\oplus A'$ for some subgroup $A'$, which can be made into a ring with $A'^2=0$. $\endgroup$
    – Jeremy Rickard
    Commented Jan 21, 2014 at 12:09
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    $\begingroup$ If $x$ has infinite order, there are some obvious restrictions on how divisible $x$ must be. The set of integers $n$ for which $ny=x$ has a solution must be closed under multiplication. $\endgroup$
    – Jeremy Rickard
    Commented Jan 21, 2014 at 12:12
  • $\begingroup$ @Jeremy Rickard: The solution for the finite exponent case is now completed by your comment. The most complicated case is probably torsion free case. $\endgroup$
    – Sh.M1972
    Commented Jan 21, 2014 at 13:05

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Maybe it will be useful for you the notion of potentially invertible elements of a semigroup introduced by E. Shutov [E.S. Lyapin, "Semigroups" , Amer. Math. Soc.].

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