# Elementary Divisor Form Uncertainty

In Goodman's Algebra: Abstract and Concrete, it proposes the theorem:

Theorem 3.6.21. (Fundamental Theorem of Finitely Generated Abelian Groups: Elementary Divisor Form). Every finite abelian group is isomorphic to a direct product of cyclic groups of prime power order. The number of cyclic groups of each order appearing in such a direct product decomposition is uniquely determined.

Then, they offer the example that the elementary divisor form of $\mathbb{Z}_{30} \times \mathbb{Z}_{50} \times \mathbb{Z}_{28}$ is $(\mathbb{Z}_{4} \times \mathbb{Z}_{2} \times \mathbb{Z}_{2}) \times \mathbb{Z}_{3} \times (\mathbb{Z}_{25} \times \mathbb{Z}_{5}) \times \mathbb{Z}_{7}$.

My question is: Is this any difference from only using prime numbers in the elementary divisor decomposition or a different ordering. For example, would it still be correct to say that the elementary divisor decomposition is $(\mathbb{Z}_{2} \times \mathbb{Z}_{2} \times \mathbb{Z}_{2} \times \mathbb{Z}_{2}) \times \mathbb{Z}_{3} \times (\mathbb{Z}_{25} \times \mathbb{Z}_{5}) \times \mathbb{Z}_{7}$ or $(\mathbb{Z}_{2} \times \mathbb{Z}_{4} \times \mathbb{Z}_{2}) \times \mathbb{Z}_{3} \times (\mathbb{Z}_{25} \times \mathbb{Z}_{5}) \times \mathbb{Z}_{7}$?

• No, it would not be correct. There is no way to replace $\mathbb{Z}_4$ by $\mathbb{Z}_2\times \mathbb{Z}_2$. Commented Mar 7, 2017 at 16:19

A product of groups does not depend on the order in which the terms are written (think of how you would prove this by constructing an isomorphism). Thus, there is no substantial difference when you swap the positions of $\mathbb Z_2$ and $\mathbb Z_4$.
However, as Dietrich Burde states in the comments to your question, you cannot replace $\mathbb Z_4$ with $\mathbb Z_2\times\mathbb Z_2$, or any other "factorization" like using a product of $n$ copies of $\mathbb Z_p$ instead of $\mathbb Z_{p^n}$ for $n\ge 2$. The two groups are actually not isomorphic. Indeed, $\mathbb Z_4$ has an element of order $4$ and $\mathbb Z_2\times\mathbb Z_2$ does not. How would you adapt that argument to prove that for all $n\ge 2$ and all prime numbers $p$, $$\mathbb Z_{p^n}\not\cong\mathbb Z_p\times \dots\times\mathbb Z_p\,\text{(n times)}?$$