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For a finite cyclic group $\mathbb{Z}/n\mathbb{Z}$ we can define an epimorphism $$\pi: \mathbb{Z}/\text{gcd}(2,n)n\mathbb{Z} \to \mathbb{Z}/n\mathbb{Z}, \quad [x] \mapsto \frac{2}{\text{gcd}(2,n)} [x].$$ I would think of $\sqrt{\mathbb{Z}/n\mathbb{Z}}=\mathbb{Z}/\text{gcd}(2,n)n\mathbb{Z}$ as the group of all squareroots of $\mathbb{Z}/n\mathbb{Z}$ (similar to a branch covering). My questions are:

  • For a general finite abelian group $G = \bigoplus_{i =1}^r\mathbb{Z}/n_i\mathbb{Z}$ we could define $\sqrt{G}$ factorwise. I don't think that this depends on the factor decomposition, but I am not sure. Does it?
  • If the above is true, is there a more abstract characterization of $\sqrt{G}$ (maybe in terms of a universal property)
  • If all of the above is true, this group probably has a name. What is it?

EDIT:

I think that $G \xrightarrow{\iota} \sqrt{G}\xrightarrow{\pi} G$ is the unique (up to isomorphism) mono-epi factorization of the square map $G \xrightarrow{2} G$ such that $G = \operatorname{ker} p\circ\pi$. Here $p:G \to G/2G$ is the canonical quotient map. In particular $2\sqrt{G}=G$.

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    $\begingroup$ * When $n$ is even, if you intend the map to be a squaring map, you don't want $[x]\mapsto[2x]$, but just $[x]\mapsto[x]$ (look at small examples). * Under this definition, your construction doesn't depend on the factor decomposition (the number of even factors is an invariant of a finite abelian group). * Note that this is not a canonical square root group! Given any square root group, you can append an arbitrary group of exponent 2 and let all those elements map to the identity. $\endgroup$
    – Greg Martin
    Commented Jul 23, 2022 at 21:43
  • $\begingroup$ Oh thx, I edited it $\endgroup$
    – Bipolar Minds
    Commented Jul 23, 2022 at 21:58
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    $\begingroup$ Please ask one question at a time. You also need context. That aside, this is an interesting question . . . $\endgroup$
    – Shaun ♦
    Commented Jul 23, 2022 at 22:35
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    $\begingroup$ Thx for your comment. It's quite late here but I will add some info tomorrow. $\endgroup$
    – Bipolar Minds
    Commented Jul 24, 2022 at 0:34
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    $\begingroup$ @MathematicalEmergency In an odd-order torsion abelian group, all elements already have square roots, so you don't have to adjoin any square roots to the group. If you imagine a cyclic group as a group of roots of unity, OP's square root of it would be the group of all square roots of elements of the original group. | BTW OP, the epimorphism is $[x]\mapsto[2x]$ if you think of it as $(\frac{1}{2}\Bbb Z)/n\Bbb Z\to\Bbb Z/n\Bbb Z$ (which connects to my previous comments by viewing residues $x$ as numerators of rational multiples $x/n$ of $2\pi i$ for phases in polar forms of complex numbers.) $\endgroup$
    – anon
    Commented Jul 24, 2022 at 0:40

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