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Let $A$, $B$, $C$ be finite abelian groups satisfying the property $$A\times B \cong A\times C$$ Does it follow that $B\cong C$?

There exists a multiplicative bijection $\phi: A\times B \rightarrow A\times C$. Let $(a, c), (a', c')\in A\times C$, then by the injectivity of $\phi$, $\phi(a, c) = \phi(a', c')$ if and only if $a = a'$, $c = c'$. By the surjectivity of $\phi$, we know for every $(a, c)\in A\times C$ there exists $(a, b)\in B$ such that $\phi(a, b) = (a, c)$. And finally, by the multiplicative property, $\phi(ac, bd) = \phi(a, b)\cdot \phi(c, d)$

But I don't see any connection that there must be an isomorphism $\mu$ between $B$ and $C$. Any hints appreciated.

Shaun
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Dee
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    Yes, this is true in general when the groups are finite. A proof may be found here. (I suggest not closing this question as a duplicate yet, incase someone types out a slick proof for the abelian case.) – user1729 May 29 '23 at 09:17
  • @user1729 There are many duplicates for every single variation of this question. Another post, with more background, is this one, but there are really many, many such posts. In particular, there are posts giving nice proofs for the fact that finite abelian groups are cancellable. – Dietrich Burde May 29 '23 at 09:21
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    @DietrichBurde I was unable to find an abelian-specific one, so thanks for digging one up! (I actually don't like the proof there - surely it can be done without reference to the FTFAG?!) – user1729 May 29 '23 at 09:30
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    @user1729 I will check the "other nine" posts on finite abelian groups, whether or not they use the structure theorem. The next one uses it more or less, too. This one, too. May be in the end, the elementary proof you have linked is better? The proof by Hirshon is often cited. – Dietrich Burde May 29 '23 at 09:33
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    @user1729 More posts about the (finitely generated) abelian case are this one and this one. The last one is by myself (surprise). It looks rather short, but I fear you still don't like it :) – Dietrich Burde May 29 '23 at 09:42
  • @DietrichBurde I was just being picky, but now I'm slightly interested. My brain is busy elsewhere this week, but maybe I'll think about a simple FTFAG-based proof when it returns. – user1729 May 29 '23 at 17:09

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