Suppose $X,Y,Z$ are finite abelian groups with $X \times Y \cong X \times Z$. How to show that $Y\cong Z$?

If we assume that we can decompose $Y,Z$ into cyclic groups that are powers of primes, I believe the problem becomes trivial. But suppose that we only know that we can decompose finite abelian groups as $\mathbb{Z}_{a_1} \times \dots \times \mathbb{Z}_{a_n}$ where $a_1|\dots |a_n$. How to proceed from there?

  • $\begingroup$ So should we take it you cannot use the classification of finite abelian groups, thus trivializing (I have no idea if this is true) the problem? $\endgroup$ – pjs36 Mar 2 '16 at 5:29
  • $\begingroup$ Yes, the only classification we were given is that one can decompose any finite abelian group into $\mathbb{Z}_{a_1} \times \dots \times \mathbb{Z}_{a_n}$ where $a_1|\dots |a_n$ (so no decomposition into prime powers is assumed, although I would suppose they are equivalent, though that is not obvious to me). $\endgroup$ – Leo Mar 2 '16 at 5:35
  • $\begingroup$ From the decomposition you were taught, it is easy to pass the the decomposition into the product of cyclic groups of prime power order. You just need to do it for each $\mathbb{Z}_{a}$, and that you do with the Chinese remainder theorem. $\endgroup$ – Andreas Caranti Mar 2 '16 at 13:20

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