I want to solve the following problem (Fuchs, "Infinite Abelian Groups", Vol.$2$, pp. $153$ Ex. $4$):

"Let $A$ be a torsion-free group of finite rank $n$ and $F$, $F'$ free subgroups of $A$ of rank $n$. Then $A/F$ and $A/F'$ are isomorphic to subgroups of $\oplus_{i=1}^n\mathbb{Q}/\mathbb{Z}$ and $A/F\oplus G\cong A/F' \oplus G' $ for suitable finite groups $G$ and $G'$."

How can I prove the existence of such groups $G$ and $G'$?

Thank you in advance for your help.

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    $\begingroup$ I am guessing that $A$ is supposed to have rank $n$. $\endgroup$ – Derek Holt Feb 16 at 13:45
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    $\begingroup$ I don't know much about abelian groups, but the idea here is that a finitely generated subgroup of $Q/Z$ is a direct sum of $C_{p^\infty}$ for finitely many primes $p$ and some finite cyclic groups. So you need to show that the primes that occur in the $C_{p^\infty}$ factors are the same in $A/F$ and $A/F'$. $\endgroup$ – Derek Holt Feb 17 at 8:19
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    $\begingroup$ As I said, I have no great expertise in abelian groups. It if not true that $A/F$ is finitely generated (the groups $Z_{p^\infty}$ are not), but I meant that its finitely generated subgroups have at most $n$ generators. $\endgroup$ – Derek Holt Feb 17 at 13:03
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    $\begingroup$ "such that" in this statement is awkward, since it relates two independent results. It should be "and". $\endgroup$ – YCor Feb 17 at 13:21
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    $\begingroup$ No, it seems the obvious natural way. $\endgroup$ – YCor Feb 17 at 13:41

The subgroup $F\cap F'$ is free abelian. Working in the quotient by this subgroup reduces to proving the following:

For an integer $n$, let $B$ be a subgroup of $(\mathbf{Q}/\mathbf{Z})^n$, and let $U,V$ be finite subgroups of $B$. Then there exist finite groups $S,T$ such that $(B/U)\times S$ is isomorphic to $(B/V)\times T$.

In an abelian group $G$, and prime $p$, let $G_p$ be the $p$-primary part (the set of element of order dividing some power of $p$). We have $B/U\simeq\bigoplus_p(B/U)_p\simeq\bigoplus_p B_p/U_p$, where the sum is over primes. For all but finitely many primes $p$, $U_p=V_p=0$. For such a prime, we have $(B/U)_p\simeq B_p\simeq (B/V)_p$. So it is enough to prove result for the finitely many individual remaining primes.

In other words, we can suppose that $B=B_p$, which is an artinian $p$-primary group. Structure of artinian abelian groups says that, for some integer $k$, $B$ is isomorphic to $C_{p^\infty}^k\oplus F_B$, where $F_B$ is a finite abelian $p$-group (namely $F_B$ is isomorphic to the quotient of $B$ by its divisible part, i.e. is the largest finite quotient of $B$). We easily deduce that $B/U$ is isomorphic to $C_{p^\infty}^k\oplus F_{B/U}$ and $B/V$ is isomorphic to $C_{p^\infty}^k\oplus F_{B/V}$. Hence $(B/U)\oplus F_{B/V}$ is isomorphic to $(B/V)\oplus F_{B/U}$.

  • $\begingroup$ I'm trying to understand your proof. Here some questions: $(1)$ why does $F\cap F'$ have rank $n$? $(2)$ $(B/U)_p=B_p/U_p$ by the $2$nd isomorphism theorem noting that $U_p=B_p\cap U$; $(3)$ I don't know artinan groups and their structure theorem. Can you give me some reference? $\endgroup$ – LBJFS Feb 17 at 18:41
  • $\begingroup$ (1) if $F\cap F'$ had rank $<n$, $F+F'$ would have rank $>n$. $\endgroup$ – YCor Feb 17 at 20:42

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