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I have two finite abelian groups and the number of elements in each with any given order is identical. Does this imply the groups are isomorphic?

geoffrey
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  • Does $\mathbb{Z}$ and $\mathbb{Z} \oplus \mathbb{Z}/2\mathbb{Z}$ suffice as a counterexample? They are non-isomorphic, and neither have elements with finite order. – AlkaKadri Aug 02 '24 at 11:26
  • @AlkaKadri Yes you are correct. However, I wanted to only consider finite groups. I edited my question. Thanks. – geoffrey Aug 02 '24 at 11:31
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    This question really is lacking sufficient context, so I am voting to close. – Derek Holt Aug 02 '24 at 11:33
  • Any finite abelian group of order $n$ is isomorphic to a product of cyclic groups with prime power order (where the primes are the prime decomposition of $n$), so yes! See https://math.stackexchange.com/questions/2729266/isomorphisms-of-finite-abelian-groups – AlkaKadri Aug 02 '24 at 11:34
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    This has been asked before, see: https://math.stackexchange.com/questions/72944/two-finite-abelian-groups-with-the-same-number-of-elements-of-any-order-are-isom

    It also appears in the form of a paper in some journal, but I don't recall the details now.

     – the_fox Aug 02 '24 at 11:38
  • I have an abelian group with 1 element of order 1, 27 elements of order 2, 8 elements of order 3, 36 elements of order 4, and 24 elements of order 6. How can I tell what group it is? – geoffrey Aug 02 '24 at 12:00
  • It does not exist. – Anne Bauval Aug 02 '24 at 12:26
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    @Anne Bauval. Yes, Thank you. – geoffrey Aug 02 '24 at 14:32

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