Section 67: Problem 6 Solution

Working problems is a crucial part of learning mathematics. No one can learn topology merely by poring over the definitions, theorems, and examples that are worked out in the text. One must work part of it out for oneself. To provide that opportunity is the purpose of the exercises.

James R. Munkres

Prove the following:

Theorem. If $A$ is a free abelian group of rank $n$ , then any subgroup $B$ of $A$ is a free abelian group of rank at most $n$ .

Proof. We can assume $A=\mathbb{Z}^{n}$ , the $n$ -fold cartesian product of $\mathbb{Z}$ with itself. Let $\pi_{i}:\mathbb{Z}^{n}\to\mathbb{Z}$ be projection on the $i$ th coordinate. Given $m\le n$ , let $B_{m}$ consist of all elements $\mathbf{x}$ of $B$ such that $\pi_{i}(\mathbf{x})=0$ for $i>m$ . Then $B_{m}$ is a subgroup of $B$ .

Consider the subgroup $\pi_{m}(B_{m})$ of $\mathbb{Z}$ . If this subgroup is nontrivial, choose $\mathbf{x}_{m}\in B_{m}$ so that $\pi_{m}(\mathbf{x}_{m})$ is a generator of this subgroup. Otherwise, set $\mathbf{x}_{m}=0$ .

(a) Show $\{\mathbf{x}_{1},\ldots,\mathbf{х}_{m}\}$ generates $B_{m}$ , for each $m$ .

(b) Show the nonzero elements of $\{\mathbf{x}_{1},\ldots,\mathbf{x}_{m}\}$ form a basis for $B_{m}$ , for each $m$ .

(a) First, we need to show that if $A$ is a subgroup of $\mathbb{Z}$ , then $A$ is either trivial or infinite cyclic. If $A$ is not trivial, then it contains a positive integer. Let $m$ be the least positive integer that belongs to $A$ . Suppose $x\in A$ . Then, there are some integer numbers $n$ and $r$ such that $x=nm+r$ and $0\le r<m$ . We have $r=x-nm$ where the right hand side is in $A$ , hence, $r$ is in $A$ . Since $m$ is the least positive integer in $A$ , $r=0$ . We conclude that for every $x\in A$ , there is some $n\in\mathbb{Z}$ such that $x=nm$ . At the same time, for every $n\in\mathbb{Z}$ , $nm\in A$ . Therefore, $A$ is generated by $\{m\}$ .

The above fact proves that $\{\mathbf{x}_{1}\}$ generates $B_{1}$ . Suppose, $\{\mathbf{x}_{1},\ldots,\mathbf{x}_{k}\}$ generates $B_{k}$ . Using the fact above, $\pi_{k+1}(\mathbf{x}_{k+1})$ generates $\pi_{k+1}(B_{k+1})$ , hence, for every $\mathbf{x}\in B_{k+1}$ , there is some $n\in\mathbb{Z}$ such that $\pi_{k+1}(n\mathbf{x}_{k+1})=\pi_{k+1}(\mathbf{x})$ . Then, $\mathbf{x}-n\mathbf{x}_{k+1}\in B_{k}$ , and can be generated by $\{\mathbf{x}_{1},\ldots,\mathbf{x}_{k}\}$ . Therefore, $\mathbf{x}$ can be generated by $\{\mathbf{x}_{1},\ldots,\mathbf{x}_{k+1}\}$ . By induction, we are done.

(b) This can be done by induction starting from the last coordinate. In fact, the step is already outlined in (a).

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Section 67: Problem 6 Solution

Section 67: Problem 6 Solution