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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.

Linearity

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