# Torsion-free Abelian Groups of Finite Rank and Free Groups (Fuchs) - Self study

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 $$G'$$?

• I am guessing that $A$ is supposed to have rank $n$. – Derek Holt Feb 16 at 13:45
• 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 $A/F'$. – Derek Holt Feb 17 at 8:19
• 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. – Derek Holt Feb 17 at 13:03
• "such that" in this statement is awkward, since it relates two independent results. It should be "and". – YCor Feb 17 at 13:21
• No, it seems the obvious natural way. – 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 $$(B/V)\oplus F_{B/U}$$.
• 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? – LBJFS Feb 17 at 18:41
• (1) if $F\cap F'$ had rank $<n$, $F+F'$ would have rank $>n$. – YCor Feb 17 at 20:42