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Pretty sure it is impossible if everything is finite, $|G_i|$ must be a prime or a power of a prime, then either LHS and RHS have different group sizes or they have different number of elements of a particular order. I just don't know about the infinity case, feels like it might not be true for some special case. Need help please.

  • $\begingroup$ any suggestion at all? $\endgroup$ Nov 7, 2014 at 14:11
  • 2
    $\begingroup$ This is relevant: there are groups which $G$ and $H$ such that $G\not\cong H$ but $G\times\mathbb{Z}\cong H\times\mathbb{Z}$. I do not know if $G$ and $H$ could be taken to be abelian. $\endgroup$
    – user1729
    Nov 10, 2014 at 12:11

1 Answer 1


This is answered in Theorem 90.1 of Infinite Abelian Groups (Vol II) by L. Fuchs, which shows that, for any $n\geq 2$, there is a finite rank abelian group which is the direct sum of two indecomposable groups and of $n$ indecomposable groups.

For the case $n=3$, one example that can be extracted from the general theorem is as follows. All the groups will be subgroups of $\mathbb{Q}^4$.

$G_1$ is generated by the elements $\left(\frac{2}{5^s},0,\frac{1}{5^s},0\right)$, $\left(0,\frac{2}{7^t},0,\frac{1}{7^t}\right)$ for all $s,t>0$ and $\left(0,0,\frac{1}{2},\frac{1}{2}\right)$.

$G_2$ is generated by the elements $\left(\frac{3}{5^s},0,\frac{1}{5^s},0\right)$, $\left(0,\frac{3}{7^t},0,\frac{1}{7^t}\right)$ for all $s,t>0$ and $\left(0,0,\frac{1}{3},\frac{1}{3}\right)$.

$G_3$ is generated by the elements $\left(\frac{1}{5^s},0,0,0\right)$ for all $s>0$.

$G_4$ is generated by the elements $\left(0,\frac{1}{7^t},0,0\right)$ for all $t>0$.

$G_5$ is generated by the elements $\left(0,0,\frac{1}{5^s},0\right)$, $\left(0,0,0,\frac{1}{7^t}\right)$ for all $s,t>0$ and $\left(0,0,\frac{1}{6},\frac{1}{6}\right)$.

It's straightforward to check that $G_1\oplus G_2=G_3\oplus G_4\oplus G_5$. Fuchs quotes a theorem from earlier in the book to prove indecomposability in the more general statement that he gives, but it's probably not too hard to check directly in this specific example.

  • $\begingroup$ is there a link for this? thanks $\endgroup$ Nov 10, 2014 at 16:23
  • $\begingroup$ I'm afraid I couldn't find it online (the book dates from the early 1970s). From memory, the construction is not enormously complicated, and I'll post it when I have time (which might not be for a couple of days, I'm afraid, so if anybody else wants to post it in the meantime, please go ahead). $\endgroup$ Nov 10, 2014 at 19:17
  • $\begingroup$ @HaipingYang I've edited my answer to give an example (the simplest special case I could see of the more general construction given in Fuchs' book). $\endgroup$ Nov 12, 2014 at 12:22

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