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The statement of the fundamental theorem of finite abelian groups I have is as follows:

Let $G$ be a finite abelian group. Then G can be written as $$\mathbb{Z}_{n_1} \oplus \cdots \oplus \mathbb{Z}_{n_s}$$ where each $n_{i+1} | n_i$ for $1 \le i \le s-1 $ and each $n_i \ge 2$. Moreover, this decomposition is unique.

I am unsure exactly what the uniqueness part means. For example, take $G$ to be a group of order 72. I can write down two compositions $$\mathbb{Z}_{24} \oplus \mathbb{Z}_{3}$$ $$\mathbb{Z}_{12} \oplus \mathbb{Z}_{6}$$ which satisfy the theorem, but do not appear to be unique, i.e. I have two distinct integer sequences $\{24, 3\}$ and $\{12, 6\}$. What am I missing?

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Those two groups are not isomorphic. It fails to be a counterexample because the theorem claims the uniqueness of a decomposition for every fixed finitely generated abelian group G. (You can clearly see that they aren’t isomorphic since the first of the two has an element of order 24 and the second one doesn’t).

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  • $\begingroup$ Of course. In fact, as a result of the theorem I know these two groups are not isomorphic. $\endgroup$
    – Paul Slevin
    Commented Nov 29, 2019 at 14:44
  • $\begingroup$ That’s right. The theorem doesn’t state that every abelian group of order 72 is isomorphic to Z_24 X Z_3 (as you’ve seen isn’t the case); but that every abelian group of order 72 is isomorphic to some direct product of cyclic groups satisfying the property you wrote. The “uniqueness” part of the theorem merely states that each of these direct products of cyclic groups (with the orders satisfying n_i+1 | n_i) are pairwise non-isomorphic. $\endgroup$
    – user622002
    Commented Nov 29, 2019 at 15:08

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