If you find any mistakes, please make a comment! Thank you.

Compute a torsion subgroup

Fix $n \in \mathbb{Z}^+$ with $n > 1$. Find the torsion subgroup of $\mathbb{Z} \times \mathbb{Z}/(n)$. Show that the set of all elements of infinite order together with the identity is not a subgroup.

Solution: We claim that the torsion subgroup is $0 \times \mathbb{Z}/(n)$.

($\subseteq$) Suppose $(a, \overline{b}) \in T$. If $a \neq 0$, then $ka \neq 0$ for all integers $k$, so that $k(a,\overline{b}) \neq 0$ for all integers $k$. Thus $$(a, \overline{b}) \in 0 \times \mathbb{Z}/(n).$$ ($\supseteq$) Note that $n (0, \overline{b}) = (0, \overline{0})$, so that $0 \times \mathbb{Z}/(n) \subseteq T$. Note that $(1,\overline{1})$ and $(-1, \overline{0})$ have infinite order, but that $$(1, \overline{1}) + (-1, \overline{0}) = (0,\overline{1})$$ is a nonidentity element of finite order. So this set is not closed under the group operator.


This website is supposed to help you study Linear Algebras. Please only read these solutions after thinking about the problems carefully. Do not just copy these solutions.
Close Menu