Let $A$ finitely generated abelian group, and $A_1 \le A$. I have to prove that $rk(A)=rk(A_1)+ rk(A/A_1)$

Question

Let be $A$ a finitely generated (f.g.) abelian group, and let be $T$ its torsion, then by the structure theorem of f.g. abelian gruops we have that $A/T \simeq \mathbb{Z}^d$, so we can define $b$ the rank of the group $A$, denoted $rk(A)$.

We say that a group $G$ (not neceassary abelian) is free if it has a free set of generators, where for "free subset" we mean a subset $X$ of $G$ such that $x_1^{\beta_1} \ldots x_n^{\beta_n}=1$, for $ x_1 \ldots x_n$ distinct elements of $X$, implies $\beta_1= \ldots =\beta_n=0$. Is easy to prove that for a f.g. abelian group the rank is the numer of elements in a free system of generators. Thinking at the structure of f.g. abelian groups we se that if $T=T(A)$ is the torsion of $A$ (f.g. abelian group) and $B \le T$, then $rk(A/B)=rk(A)$.

For $A$ f.g. abelian gruop are equivalent:

1. $A\simeq \mathbb{Z}^d$.
2. the torsion subgroup $T(G)$ of $G$ is trivial.
3. $A$ is free.

Fact1: If $\{ x_1 \ldots x_n \}$ is a free system of generators for a f.g. abelian group $A$, then $rk(A)=n$ and if $A_1=\langle x_1 \ldots x_d\rangle$, with $d \le n$, then $\{ x_1 \ldots x_d \}$ is free set of generators of $A_1$ which have rank $d$. It is easy to prove that $\{ x_{d+1}+A_1 \ldots x_n+A_1 \}$ is free set of generators of $A/A_1$, so in this case we observe that hold $rk(A)=rk(A_1)+rk(A_2)$ with $A_2=A/A_1$.

Now my problem is generalize to a generic f.g. abelian group $A$ and to $A_1$ generic subgroup of $A$, showing that $rk(A)=rk(A_1)+rk(A_2)$ holds, with $A_2=A/A_1$ as before.

First I observed that we can suppose that $A$ is free, replacing $A$ with $A/T$ (T is the torsion of $A$) and $A_1$ with $(A_1+T)/T$. Indeed $rk((A_1+T)/T)=rk(A_1)$, $rk(A/T)=rk(A)$ and because $(A_1+T)/T \le T(A/A_1)$ for our first observation, named $B=(A_1+T)/T$ and $C=(A_1+T)/A_1$, we have that $rk((A/T)/B)=rk(A_2/C)=rk(A_2)$. Now suppose that $A$ is free abelian f.g, then is direct sum of $n=rk(A)$ copies of $\mathbb{Z}$. Taken $A_1 \le A$, is well known (see theorem 10.17 of Introduction of Theory of Groups, Rotman) that, since $A$ is abelian f.g., $A_1$ is free f.g.

Now if we can prove that:

Fact2: exists for $A$ a free system of generators $\{x_1 \ldots x_n\}$ such that $A_1=\langle m_1 x_1 \rangle \times \ldots \times \langle m_k x_k \rangle$ with $k \le n$ (the $\times$ mean, of course, the internal DIRECT product), for suitable choose of $m_i \in \mathbb{N} \setminus \{0\}$

then taken $I=\langle x_1 \rangle \times \ldots \times \langle x_k \rangle$ we have that obviously $rk(A_1)=rk(B)=k$ and because $B/A_1 \le T(A/A_1)$ for our first observations and then we have that $rk(A/A_1)=rk(A/B)=n-k$ and we conclude for fact2.

In this way the problem is reduced to prove the fact2. I have no idea to prove it, I also found this page where the last theorem stated is similar to fact2, but the generators for the subgroup not need to be free, and I don't know how to get a free set generators from a generic set of generators because for a null $\mathbb{Z}$-linear combination $0=\sum m_i x_i$ (the $m_i$ are integers) I can't "isolate" one generator $x_i$, since $\mathbb{Z}$ isn't a field.

Thank you.

Answer 1

The rank of a finitely generated abelian group $A$ is the dimension of $A\otimes_{\mathbb{Z}}\mathbb{Q}$ as a vector space over $\mathbb{Q}$. Since $\mathbb{Q}$ is flat as a $\mathbb{Z}$-module, from the exact sequence $$ 0\to A_1\to A\to A/A_1\to 0 $$ you get the exact sequence $$ 0\to A_1\otimes_{\mathbb{Z}}\mathbb{Q} \to A\otimes_{\mathbb{Z}}\mathbb{Q} \to (A/A_1)\otimes_{\mathbb{Z}}\mathbb{Q}\to 0 $$ and you're done.