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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?

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  • $\begingroup$ 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
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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

$$\mathbb{Z}^n\oplus\mathbb{Z}_{p^{k_1}}\oplus\cdots\oplus\mathbb{Z}_{p^{k_m}}$$

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

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

i.e.

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

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.

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  • $\begingroup$ 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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