If A, B, C be abelian grops and if A isomorph with direct sum of B and C and A be isomorph with B what we can say about C?
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$\begingroup$ @Rahman: that's a fancy way to ask about $B$ being isomorphic to $B+C$. (Especially in the view of the two answers below :-). $\endgroup$– Włodzimierz HolsztyńskiCommented Jun 23, 2013 at 17:50
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1$\begingroup$ More interesting question would be: Given $C$, what can we say about groups $B$ such that the direct sum $B+C$ is isomorphic to $B$? $\endgroup$– Włodzimierz HolsztyńskiCommented Jun 23, 2013 at 17:57
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2$\begingroup$ The post by guest correctly answers the question. This question deserves to be closed now. Please, no changing the question, reinterpreting the question, changing hypotheses, etc.: let OP ask a new question elsewhere if something better comes to mind. $\endgroup$– Todd TrimbleCommented Jun 23, 2013 at 21:16
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$\begingroup$ +1 Todd. Stone soup, and all that $\endgroup$– Yemon ChoiCommented Jun 23, 2013 at 21:24
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3 Answers
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If $A$ is finitely generated then $C$ is necessary trivial by the fundamental theorem of finitely generated abelian groups. In the case in which $A$ is not finitely generated, I like in particular the following counterexample (which you can find in the Isaac`s book ``Algebra. A graduate course"): One can see that $({\mathbb R},+)$ is isomorphic to the direct sum of two copy of itself but, of course, $({\mathbb R},+)$ is not trivial... Anyway, this is not a research level question!
answered Jun 23, 2013 at 17:19
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Nothing, let A be the direct sum of infinite copies of C. Then A is isomorphic to A direct sum C.
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$\begingroup$ For that matter, let A be B direct summed with countably (or more) many copies of C. I think under certain moderate conditions this can characterize such A, but I don't have chapter 5 of "Algebras, Lattices, Varieties" handy. Gerhard "Ask Me About System Design" Paseman, 2013.06.23 $\endgroup$ Commented Jun 23, 2013 at 18:51
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The question is also related to the so called cancellation problem for (not necessarily abelian) groups:
If $G\times A\cong G\times B$, when does it follow that $A\cong B$ ?
Perhaps it is worth to note that there is a positive answer for finite groups,
and in some other related cases (if $G$ is to-Hom finite, and $A$ and $B$ are finite).
Examples of to-Hom groups are quasicyclic groups and torsion-free groups.
It is not true in general, of course, with the counterexample given in the above answer, where $A$ is a nontrivial group and $G$ the countable direct product of $A$ with itself. Then $G\times A \cong G \times \lbrace e \rbrace$, but $A\neq \lbrace e \rbrace$.
answered Jun 23, 2013 at 20:07