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I was thinking that the structure theorem for abelian groups is not true for the trivial group. Am I missing something or is the theorem just valid for groups of order $>{1}$ ?

Edit: The theorem I'm working with is: Let $G$ be an abelian group of finite order, then there exist integers $d_1,...,d_k$ so that $$G\sim\Bbb{Z}/(d_1)\times...\times\Bbb{Z}/(d_k),$$ and so that $d_i\mid{d_{i+1}}$ for $1\leq{i}\leq{k-1}$. The trivial group clearly doesn't satisfy this.

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  • $\begingroup$ I'm not sure what form you were given the structure theorem in, but you can see the trivial group as a trivial product: no finite factors, times $\mathbb{Z}$ to the power 0. $\endgroup$
    – Chessanator
    Commented Jan 15, 2018 at 15:48
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    $\begingroup$ Another way: $\mathbb Z / (1)$ is the trivial group. $\endgroup$
    – lisyarus
    Commented Jan 15, 2018 at 17:23

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I'm guessing you're talking about the structure theorem for finitely generated abelian groups, which is a product of groups of some form (depending on the version of the theorem you're talking about.)

The interpretation that works (which is perfectly sound) is that $\{0\}$ is the empty product of abelian groups.

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