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$A,B,C$ are groups. Suppose that $A \times B \cong A \times C$ and both max-$n$ and min-$n$ hold in these groups. Prove that $ B \cong C$ (so that $A$ may be "cancelled"). Show that this condition is not generally valid.

The conditions max-$n$ and min-$n$ refer to the maximal and minimal conditions on normal subgroups respectively.

Let $\phi: A \times B \rightarrow A \times C$ be an isomorphism between the two product groups. I am considering to construct an automorphism of $A \times C$, which maps $\phi(1 \times B)$ onto $1 \times C$, or an automorphism of $A \times B$, mapping $1 \times B$ onto $\phi^{-1} (1 \times C)$. But I don't know how to apply the conditions...

Thank you very much!

Alejandro Tolcachier
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ShinyaSakai
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    Sorry, are these like the ascending/descending chain conditions for groups? – Dylan Moreland Aug 19 '11 at 16:08
  • This is the simplest case of the Krull-Schmidt Theorem. Usually one establishes some properties of some special kinds of endomorphisms called "normal endomorphisms" (endomorphisms such that $af(b)a^{-1} = f(aba^{-1})$. (There are other ways, but this is common; see e.g., Hungerford). What kind of results have you established for groups with chain conditions on normal subgroups? – Arturo Magidin Aug 19 '11 at 16:27
  • Are you saying $A,B, C,$ and $A\times B$ all have the min-n and max-n property or just $A\times B?$ – jspecter Aug 19 '11 at 18:25
  • In fact, you only need $A$ to have min-n (no conditions on $B$ or $C$) in order to cancel (this is non-trivial). For an example where cancellation fails when $A$ doesn't have min-n, and a reference to relevant article, see http://math.stackexchange.com/questions/32610/non-isomorphic-groups-who-product-with-z-is-isomorphic/32614#32614. –  Aug 19 '11 at 20:41
  • @Dylan Moreland: Thank you for your comment. Yes, max-$n$ and min-$n$ mean the ascending and descending chain conditions on normal subgroups. – ShinyaSakai Aug 20 '11 at 17:55
  • @jspecter: Thank you for your comment. I have copied the problem from a text book. To my understanding, literally, the writer supposed all the groups to satisfy the conditions. But as Steve D has pointed out, the condition could be weaker. – ShinyaSakai Aug 20 '11 at 18:09
  • @Arturo Magidin: Thank you very much. I didn't find that Krull-Schmidt Theorem could be applied. – ShinyaSakai Aug 20 '11 at 18:19
  • @Steve D: Thank you very much. I see the example: $A$ is the free abelian group with countably many generators, and $B$ is the free abelian group with generated by only $1$ element. $C$ is trivial. Then $A \times B \cong A \times C$, but $B \not \cong C$. http://www.artofproblemsolving.com/Forum/viewtopic.php?f=61&t=351637 – ShinyaSakai Aug 20 '11 at 18:22
  • @Shinya: I did not thing that you could apply Krull-Schmidt. Rather, my point is that this is essentially the base case in the induction for the uniqueness clause of the proof of the Krull-Schmidt Theorem. The ways in which I know how to prove this result all require some of the machinery that is built up to prove the K-S Theorem (though of course you do not need the full K-S to prove this). Hence my question as to what auxiliary results you might know, vs. what needs to be developed. – Arturo Magidin Aug 20 '11 at 20:45

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