Additivity of Rank of an Abelian group

Question

I am studying algebraic topology from Rotman and I needed to use rank of an (not necessarily free) Abelian group and in the exercises I faced a problem saying, given an exact sequence $0 \rightarrow A \rightarrow B\rightarrow C\rightarrow 0$ we have $rank A+ rank C=rank B$. The definition of rank given in the book is that:

$\mathbf{Definition:}$ For an Abelian group G, we define $rank(G)=r$ if G has a free Abelian Subgroup F of rank $r$ such that $G/F$ is torsion.

Then it describes that taking a maximal independent subset (i.e., a subset $B \subset G$ such that $\sum m_bb=0$ With $m_b\in \Bbb Z$ and $m_b=0$ a.e. $\Rightarrow m_b=0$ $\forall b$), if we take the Subgroup generated by it (which is necessarily free) it will suffice as F.

The problem stated above comes with a hint that take a maximal independent subset of A and extend it to a maximal independent subset of B. I couldn’t follow this. If I try to follow the method in vector spaces I am missing the fact that the kernel was free in that case. Any help in this would be appreciated.

I proved this statement using some different idea.

Firstly, the definition of rank is independent of choice of $F$ since if we apply tensoring by $\Bbb Q$ on the sequence $0\rightarrow F \rightarrow G \rightarrow G/F \rightarrow 0$ then we get using $\Bbb Q \otimes G/F=0 $ that, $dim_{\Bbb Q} (\Bbb Q \otimes F)=rank(F)=dim(\Bbb Q \otimes G)$ and by welldefinedness of rank of a free module we arrive at the welldefinedness of $rank(G)$.

So, tensoring the exact sequence $0 \rightarrow A \rightarrow B\rightarrow C\rightarrow 0$ by $\Bbb Q$ We get the result from the additivity of dimension of vector spaces.

I would like to know whether my approach is correct and also how can I proceed along the line of the hint given there.

(Edit: I have added a possible way out for my query. Now I am willing to know whether taking a Maximal independent subset S of A, pushing it to B and then extending the resulting set to a maximal independent subset R of B $\Rightarrow R-S $ is a maximal independent subset of C?)

Answer 1

Well along the line of the hints given, I could figure out one possible solution: We can take A maximal independent set of A push it to B by the injection given. Further choose a maximal independent set of C and pull it back to B using the given surjection. The union of these two will be a maximal independent subset of B.

(Edit: For the edited part of my question, Let $f: A \rightarrow B$ and $g:B \rightarrow C$ be the given maps. Then suppose $R$ is a maximal independent subset of A then $f(R)$ is independent subset of B extending it to $S=f(R) \cup T$ a maximal independent subset of B.

Our claim is to show that $g(T)$ is maximal independent subset of C. Suppose, $\sum (x_t.g(t))=0$ then $\sum x_t.t \in ker(g)=im(f)$ so, there is a $b \in A$ such that $f(b)= \sum(x_t.t)$ now by maximality of $R$, there is a $x>0$ Integer such that $xb=\sum(x_r.r)$, applying $f$ and by independence of $S$ We get that $g(T)$ is independent.

For maximality, let $c\in C-g(T)$ with $c=g(b)$ Then, there is $z>0$ integer such that $zb=\sum(x_r.f(r))+ \sum(x_t.t)$, applying, $g$ we get the desired result.)

Answer 2

Let $\{a_1,...,a_k\}$ be a maximal independent subset of $A$, let $\{b_1,...,b_r\}$ be a maximal independent subset of $B$, and let $f:A \hookrightarrow B$ and $g:B \twoheadrightarrow C$ be the homoprhisms in the exact sequence assumption.

The point is to consider the free groups $A' = \text{span} \{a_1,...,a_k\}$, $C' = \text{span} \{c_1,...,c_r\}$, and a free subgroup of $B$ that is equal to $f(A'')+ g^{-1}(C'')$ for some $A''\subset A$ and $C''\subset C$.

Lemma 1 The set $\{f(a_1),...,f(a_k)\}$ is linearly independent and the function $f':A' \rightarrow B'$ defined by $f' = f|_{A'}$ is an injective homorphism.

(Proof): Suppose it were not, then we would have $\sum_i n_if(a_i)=0$ for non-trivial $n_1,...,n_k$ but then $f(\sum_i n_ia_i) = 0$. Since $\{a_1,...,a_k\}$ is linearly independent we must have that $\sum_i n_ia_i \neq 0$ but this contradict the infectivity of $f$. The second statement is trivial since it inherits all of those properties from $f$. QED

Use the axiom of choice to construct a set of representatives $\{g^{-1}(c_1),...,g^{-1}(c_r)\}$; then we have the following:

Lemma 2 The set $\{g^{-1}(c_1),...,g^{-1}(c_r)\}$ is linearly independent, $\text{span} \{g^{-1}(c_1),...,g^{-1}(c_r)\}$ is a free group, and the function $g':g^{-1}(C') \rightarrow C'$ defined by $g' = g|_{g^{-1}(C')}$ is a surjective homorphism.

(Proof): Suppose it were not, then we would have $\sum_i n_ig^{-1}(c_i)=0$ for non-trivial $n_1,...,n_r$ but then $g(\sum_i n_ig^{-1}(c_i)) = g(g^{-1}(\sum_i n_ic_i)) = \sum_i n_ic_i = 0$, which contradicts the fact that $ \{c_1,...,c_r\}$ is linearly independent. The fact that $\{g^{-1}(c_1),...,g^{-1}(c_r)\}$ is linearly independent implies that $\text{span} \{g^{-1}(c_1),...,g^{-1}(c_r)\}$ is a free group. The last statement is trivial since it inherits all of those properties from $g$. QED

The last step is to consider the subgroup $C' = \text{span} \{f(a_1),...,f(a_k)\} + \text{span} \{g^{-1}(c_1),...,g^{-1}(c_r)\}$ which has the following nice property

Lemma 3 The sequence $0 \rightarrow A' \rightarrow B' \rightarrow C' \rightarrow 0 $ is exact. Furthermore we have that $\text{rank}(A') + \text{rank}(C') = \text{rank}(B') $.

(Proof): The exactness of $f',g'$ follows from the exactness of $f,g$ and the second statement is equivalent to the rank-nullity theorem after tensoring by $\mathbb{Q}$. In other words, apply rank-nullity to the vector spaces \begin{equation} 0 \rightarrow \mathbb{Q} \otimes A' \rightarrow \mathbb{Q} \otimes B' \rightarrow \mathbb{Q} \otimes C' \rightarrow 0 . \end{equation} QED

The final step is to prove that:

Lemma 4 The set $ \{f(a_1),...,f(a_k), g^{-1}(c_1),...,g^{-1}(c_r)\} $ is a maximally linearly independent subset of $B$.

(Proof): To prove that it is linearly independent notice that it is enough to consider the sums $\sum_in_i f(a_i) + \sum_jm_j g^{-1}(c_j) $ where both $n_i$ and $m_j$ are non-trivial combinations since we already proved that there are individually linearly independant. If $ \{f(a_1),...,f(a_k), g^{-1}(c_1),...,g^{-1}(c_r)\} $ was not linearly independent then we would have that $\sum_in_i f(a_i) = \sum_jm_j g^{-1}(c_j) $ and therefore, since $\text{image}(f') \subset \text{kernel} (g')$, we have that $0=g(\sum_in_i f(a_i)) = g(\sum_jm_j g^{-1}(c_j)) = \sum_j m_jc_j $ which contradicts the linear independence of the $c_i$. To see that this linearly independent set cannot be made bigger notice that $ \{f(a_1),...,f(a_k), g^{-1}(c_1),...,g^{-1}(c_r),b\} $ was linearly independent then we would have that $\text{image}(f') \neq \text{kernel} (g')$ which would contradict the exactness of $f',g'$.

Corollary We have that $\text{rank}(A) + \text{rank}(C) = \text{rank}(B) $.

I wish you the best on your future mathematical voyages! Good luck! Bon Voyage!