Section 67: Problem 4 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

The order of an element $a$ of an abelian group $G$ is the smallest positive integer $m$ such that $ma=0$ , if such exists; otherwise, the order of $a$ is said to be infinite. The order of $a$ thus equals the order of the subgroup generated by $a$ .

(a) Show the elements of finite order in $G$ form a subgroup of $G$ , called its torsion subgroup.

(b) Show that if $G$ is free abelian, it has no elements of finite order.

(c) Show the additive group of rationals has no elements of finite order, but is not free abelian. [Hint: If $\{a_{\alpha}\}$ is a basis, express $\tfrac{1}{2}a_{\alpha}$ in terms of this basis.]

According to the definition, $0$ has order $1$ . Hence, in (b) it is assumed that the element is not $0$ .

(a) If $a$ and $b$ have finite orders, then $ma=0=nb$ for some $m,n>0$ , and $\mbox{lcm}(m,n)(a+b)=0$ .

(b) If $ma=0$ for some $m>0$ , then let $a=\sum_{\alpha\in J}m_{\alpha}a_{\alpha}$ , then $0=ma=\sum_{\alpha\in J}mm_{\alpha}a_{\alpha}$ , which in the free abelian group implies that $mm_{\alpha}=0$ for all $\alpha$ , hence, $a=0$ . So, the only element of $G$ that has a finite order is $0$ .

(c) If $mz=0$ for $m>0$ , then $z=0$ . Further, using the hint, $\tfrac{1}{2}a_{\beta}=\sum_{\alpha\in J}m_{\alpha}a_{\alpha}$ implies $a_{\beta}=\sum_{\alpha\in J}2m_{\alpha}a_{\alpha}$ implies $2m_{\beta}=1$ , which is not possible.

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Section 67: Problem 4 Solution

Section 67: Problem 4 Solution