Let $A,B,C$ be finitely generated abelian groups. I want to show that $$A\oplus C \cong B \oplus C $$ then $A \cong B$. My idea is as follows: we begin by noting that this has already been proved for $A,B,C$ finite. Let

$A = \mathbb{Z}^{k_1}\oplus A'$

$B = \mathbb{Z}^{k_2}\oplus B'$

$C = \mathbb{Z}^{k_3}\oplus C'$

Then we have $$\mathbb{Z}^{k_1+k_3}\oplus A'\oplus C' \cong \mathbb{Z}^{k_2+k_3}\oplus B'\oplus C'$$ Therefore $k_1 = k_2$ and then I would like to divide out by the infinitary parts to get $$A'\oplus C'\cong B'\oplus C'$$ and since all groups are finite we can use previous theorems to derive $A'\cong B'$ hence $A\cong B$. However I am not sure how to get past the 'dividing bit' as to assume that we can cancel would be circular. How should I proceed? Is the assumption that finitely generated abelian groups cancel even true?

  • $\begingroup$ What "previous theorems"? So far you reduced the problem to showing that finite abelian groups have cancellation property (and even this step is fishy, how did you deduce that $k_1=k_2$?). Anyway the statement quite simply follows from the classification of finitely generated abelian groups. Since the decomposition of a finitely generated abelian group into product of $\mathbb{Z}$'s and $\mathbb{Z}_{p^k}$'s is unique. $\endgroup$ – freakish Apr 19 '18 at 14:48
  • $\begingroup$ Previous theorem is 'Finite abelian groups have the cancellation property'. I deduce that $k_1 = k_2$ because isomorphic groups must have the same rank. Deducing the theorem from the classification of finitely generated abelian groups is precisely what this question is trying to formalize. $\endgroup$ – Elie Bergman Apr 19 '18 at 14:51
  • $\begingroup$ That is a more complicated case. Your answer to math.stackexchange.com/questions/2186770/… seems to provide a method that would work in this case. I am not interested in non-abelian groups at the moment. $\endgroup$ – Elie Bergman Apr 19 '18 at 14:56
  • $\begingroup$ Yes, for finitely generated abelian groups this is better, I agree. $\endgroup$ – Dietrich Burde Apr 19 '18 at 15:00

This follows from the classification of finitely generate abelian groups, more precisely from the fact that every finitely generated abelian group can be written as


and this decomposition is unique.

So assume that $A\oplus C\simeq B\oplus C$. Decompose each group:

$$A=\mathbb{Z}^{n_A}\oplus(\mathbb{Z}_{p^{*}})$$ $$B=\mathbb{Z}^{n_B}\oplus(\mathbb{Z}_{q^{*}})$$ $$C=\mathbb{Z}^{n_C}\oplus(\mathbb{Z}_{r^{*}})$$

I've obviously simplified the right side, they can have multiple elements. Anyway we have

$$\mathbb{Z}^{n_A}\oplus(\mathbb{Z}_{p^{*}})\oplus \mathbb{Z}^{n_C}\oplus(\mathbb{Z}_{r^{*}})\simeq\mathbb{Z}^{n_B}\oplus(\mathbb{Z}_{q^{*}})\oplus \mathbb{Z}^{n_C}\oplus(\mathbb{Z}_{r^{*}})$$


$$\mathbb{Z}^{n_A+n_C}\oplus(\mathbb{Z}_{p^{*}}\oplus\mathbb{Z}_{r^{*}})\simeq \mathbb{Z}^{n_B+n_C}\oplus(\mathbb{Z}_{q^{*}}\oplus\mathbb{Z}_{r^{*}})$$

The uniqueness kicks in and gives us $n_A+n_C=n_B+n_C$ hence $n_A=n_B$. Analogously the uniqueness implies $(\mathbb{Z}_{p^*})\simeq(\mathbb{Z}_{q^*})$ for the finite case.

  • $\begingroup$ Am I right that if $A,B,C$ are finitely generated then $A\oplus C$ and $B\oplus C$ is also finitely generated? I guess that you are using this fact in your solution $\endgroup$ – ZFR Nov 23 '18 at 23:11
  • $\begingroup$ @ZFR I'm not using that here, but yeah, finite product of finitely generated groups is finitely generated. $\endgroup$ – freakish Nov 23 '18 at 23:20

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