Exercise 2.1.7 (Torsion subgroup of $\mathbb{Z} imes (\mathbb{Z}/n \mathbb{Z})$)

Question

Fix some $n \in \mathbb{Z}$ with $n > 1$. Find the torsion subgroup (cf. the previous exercise) of $\mathbb{Z} 	imes (\mathbb{Z}/n \mathbb{Z})$. Show that the set of elements of infinite order together with the identity is {\bf not} a subgroup of this direct product.

Richard Ganaye · Accepted Answer

ewcommand{\Z}{\mathbb{Z}}

ewcommand{\N}{\mathbb{N}}

\begin{proof} 
Consider the group $G = \mathbb{Z} 	imes (\mathbb{Z}/n \mathbb{Z})$, where $0_G = (0,\overline{0})$ is the identity element.

Let $(x, \overline{y})$ be any element in $G = \mathbb{Z} 	imes (\mathbb{Z}/n \mathbb{Z})$.
\begin{itemize}
\item If $x 
e 0$, then for all $k \in \Z^+$,
$$k(x,\overline{y}) = (kx, n\overline{y}) 
e (0,\overline{0}),$$
since $kx 
e 0$.
\item If $x = 0$, then 
$$n(x,\overline{y}) = (nx, n\overline{y}) = (0,\overline{0}).$$
so the order of $(x,\overline{y})$ is finite (it divides $n$).
\end{itemize}
This shows that the torsion subgroup of $G$ is $T = \{0\} 	imes \Z/n\Z$.

\bigskip
Consider the set $S = (G \setminus T) \cup \{(0, \overline{0})\}$ (the set of elements of infinite order together with the identity).

Note that $u = (2,\overline{1}) \in S$ and $v = (-2 ,\overline{0}) \in S$ (the order of $u$ and $v$ is $\infty$), but 
$$u+v = (0 , \overline{1}) 
ot \in S,$$
since  the order of $u+v$ is finite, and $\overline{1} 
e \overline{0}$ (because $n>1$), so $u+v 
e 0_G$.

This shows that the set of elements of infinite order together with the identity is not a subgroup of $G$.
\end{proof}

Exercise 2.1.7 (Torsion subgroup of $\mathbb{Z} \times (\mathbb{Z}/n \mathbb{Z})$)

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Exercise 2.1.7 (Torsion subgroup of $\mathbb{Z} \times (\mathbb{Z}/n \mathbb{Z})$)

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