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Let $G,H,$ and $K$ be finite abelian groups. If $G \times H \cong G\times K$ then $H\cong K$.

I am trying to use the fundamental theorem for abelian groups to solve this, it is clear intuitively that I can decompose $G \times H$ and $G \times K$ and then cancel out the factors of $G$ but I have no clue on writing this rigorously, or if my method is right.

Edit: Can I say that $H \cong \Bbb{Z}_{p^i} \times \ldots \times \Bbb{Z}_{p^n}$ and $K \cong \Bbb{Z}_{p^q} \times \ldots \times \Bbb{Z}_{p^k}$ and since $G \times H \cong G\times K$ then $G\times \Bbb{Z}_{p^i} \times \ldots \times \Bbb{Z}_{p^n} \cong G \times \Bbb{Z}_{p^q} \times \ldots \times \Bbb{Z}_{p^k}$ , but the representation is unique hence $\Bbb{Z}_{p^i} \times \ldots \times \Bbb{Z}_{p^n} \cong \Bbb{Z}_{p^q} \times \ldots \times \Bbb{Z}_{p^k}$ that is $H \cong K$?

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    $\begingroup$ The mentioned theorem has a part about the uniqueness of the decomposition. Try to use this! $\endgroup$ – user26857 Apr 23 '13 at 21:41
  • $\begingroup$ This is contained in this post; also look at this post. $\endgroup$ – Zev Chonoles Apr 23 '13 at 21:58
  • $\begingroup$ @ZevChonoles I saw both of these and I found them a bit advanced and a bit unclear, hence I asked this question. $\endgroup$ – user10444 Apr 23 '13 at 22:03
  • $\begingroup$ @user10444: related generalization math.stackexchange.com/questions/20664 $\endgroup$ – Watson Aug 22 '16 at 13:40
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Recall that the Fundamental Theorem for finitely generated Abelian groups (which applies of course to finite abelian groups) states that every finite abelian group is isomorphic to, and can be decomposed uniquely as, the direct product of cyclic groups of prime order and/or cyclic groups of order equal to the power of a prime.

So each of $G, H, K$ can be decomposed uniquely in such a manner, and so the direct product of $G\times H$ and of $G\times K$ can each thereby be expressed uniquely (up to isomorphism) as the direct product of cyclic groups.

  • Since $G$ is a common factor to each of $G \times H$ and $G\times K$, and their decompositions are unique up to isomrophism,
  • and given $G\times H \cong G\times K$

then what does this imply about the relationship between $H$ and $K$?

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  • $\begingroup$ I tried using the hint and edited it into my question I still feel there is something wrong though. $\endgroup$ – user10444 Apr 23 '13 at 22:26
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    $\begingroup$ Yes, since the representation is unique up to isomorphism, the decomposition of $H$ must be isomorphic to $K$, since $G \times H \cong G \times K$. I.e., whatever the representation of H, it varies only from that of $K$ in terms of the arrangement of its factors (arrangement of factors does not affect uniqueness: one arrangement is equivalent to another, so long as each and only those $\mathbb Z p_i^k$ that appear in one appear in the other): hence the uniqueness of $G\times H \cong G\times K$ guarantees the uniqueness of $H \cong K$, up to isomorphism. $\endgroup$ – amWhy Apr 23 '13 at 22:36
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    $\begingroup$ Yes, meaning, your argument follows as stated. $\Bbb{Z}_{p^i} \times \ldots \times \Bbb{Z}_{p^n} \cong \Bbb{Z}_{p^q} \times \ldots \times \Bbb{Z}_{p^k}$...means that the number of factors $\Bbb Z_{p^j}$ in each is the same, and the particular $p$'s and particular $j$'s in the $p_j$'s of $H, K$ can be matched one-to-one, though they may appear in a different order. $\endgroup$ – amWhy Apr 23 '13 at 22:41
  • $\begingroup$ Thank you very much! I understood the whole uniqueness thing in a wrong way. $\endgroup$ – user10444 Apr 23 '13 at 22:45
  • $\begingroup$ You're welcome! Glad to help. $\endgroup$ – amWhy Apr 23 '13 at 22:48

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