2
$\begingroup$

Let $F$ be a free abelian group (of finite rank). For every two direct summands $A$ and $B$ of $F$, either $A$ is a direct summand of $B$ or $B$ is a direct summand of $A$ up to isomorphism.

My question is:

Which abelian groups have this property other than free abelian groups?

$\endgroup$
0
2
$\begingroup$

If $G$ is a finitely generated abelian group then the classification of finitely generated abelian groups can be used to classify all such groups.

Write

$$G\simeq\mathbb{Z}^n\oplus F$$

where $F$ is finite. If $F$ is non-trivial (i.e. $G$ is not free) and $n>0$ then obviously this group cannot have this property since $\mathbb{Z}^n$ cannot be a direct summand of $F$ and vice versa.

So we are left with the case when $G$ is finite. If $|G|$ is divisible by two different primes then $G\simeq H\oplus K$ with $\gcd(|H|,|K|)=1$ and hence neither $H$ can be a summand of $K$ nor vice versa.

Thus we are left with $G$ being a $p$-group. You can write $G$ as

$$G\simeq\mathbb{Z}_{p^{a_1}}\oplus\cdots\oplus\mathbb{Z}_{p^{a_m}}$$

Now every component is indecomposable and thus it can be easily seen that (by the uniquness of decomposition) that the only way for $G$ to have the desired property is when $a_1=\cdots=a_m$. And indeed every such group has the desired property.

I cannot help you in general abelian case.

$\endgroup$
1
  • $\begingroup$ Thank you very much for your exact explanation. $\endgroup$
    – M.Ramana
    May 10 '18 at 14:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.