Exercise 2.1.6 (Torsion subgroup of $G$)

Proof.

(a)

Let $G$ be an abelian group with identity element $e$, and $$T=\{g\in G\mid \exists <!--\; FUNCTION\; APPLICATION\; -->n\in {\mathbb{Z}}^{\text{+}},g^n=e\}$$

the set of elements of $G$ of finite order.

• Since $e^1=e$, $e\in T$, so $T\ne \varnothing$.

• If $g,h\in T$, then there are positive integers $n,m$ such that $g^n=e$ and $h^m=e$. Since $G$ is an abelian group,

  $${(g{h}^{-1})}^{\mathit{nm}}=g^{\mathit{nm}}{(h^{-1})}^{\mathit{nm}}={(g^n)}^m{(h^m)}^{-n}=e,$$

  thus $g{h}^{-1}\in T$

When $G$ is abelian, $T$ is a subgroup of $T$.

(b)

Consider the two transformations of $\mathbb{Z}$ given by $$t\{\begin{array}{ccc}\hfill \mathbb{Z}\hfill & \hfill \to \hfill & \mathbb{Z}\hfill \\ \hfill x\hfill & \hfill \mapsto \hfill & x+1\hfill \end{array},s\{\begin{array}{ccc}\hfill \mathbb{Z}\hfill & \hfill \to \hfill & \mathbb{Z}\hfill \\ \hfill x\hfill & \hfill \mapsto \hfill & -x\hfill \end{array},$$

and let $D_{\infty }=⟨s,t⟩$ the subgroup of $S_{\mathbb{Z}}$ generated by $s$ and $t$. Note that for all $x\in \mathbb{Z}$,

$$\begin{array}{llll}\hfill (\mathit{ts})(x)& =t(s(x))=-x+1,& \hfill & \\ \hfill (s{t}^{-1})(x)& =-(x-1)=-x+1,& \hfill & \end{array}$$

so $s{t}^{-1}=\mathit{ts}$ (therefore $\mathit{st}=t^{-1}s$), and $s^2=e$. Hence every element $g$ of $D_{\infty }$ is of the form

$$g=t^h{s}^k,h\in \mathbb{Z},k\in \{0,1\}:$$

$$D_{\infty }=\left\{t^h{s}^k\mid h\in \mathbb{Z},k\in \{0,1\}\right\}$$

Since $|s|=2$, and $|t|=\infty$, these elements are distinct.

We show by induction that $s{t}^k=t^{-k}s$ for all $k\in \mathbb{N}$. First $s{t}^0=t^{-0}s$. If $s{t}^k=t^{-k}s$ for some $k\in \mathbb{N}$, then

$$s{t}^{k+1}=(s{t}^k)t=t^{-k}\mathit{st}=t^{-k}{t}^{-1}s=t^{-k-1}s,$$

and the induction is done. So $s{t}^k=t^{-k}s$ is true for all $k\in \mathbb{N}$, therefore $s{t}^{-k}=t^{k}s$. This shows that

$$\forall <!--\; FUNCTION\; APPLICATION\; -->k\in \mathbb{Z},s{t}^k=t^{-k}s.$$

Since $|t|=\infty$, $|t^k|=\infty$. Moreover,

$${(t^{k}s)}^2=t^{k}s{t}^{k}s=t^k{t}^{-k}\mathit{ss}=e,$$

so $|t^{k}s|=2$ for all $k\in \mathbb{Z}$. Hence the set $T=\{g\in D_{\infty }\mid \exists <!--\; FUNCTION\; APPLICATION\; -->n\in {\mathbb{Z}}^{\text{+}},g^n=e\}$ is the set of elements $t^{k}s(k\in \mathbb{Z})$ of order $1$ or $2$.

Then $u=\mathit{ts}\in T$ and $v=t^{-1}s\in T$, but $\mathit{uv}=\mathit{ts}{t}^{-1}s=t^2\notin T$, so $T$ is not a subgroup of $D_{\infty }$.

The set of elements of finite order in $D_{\infty }$ is not a subgroup of $D_{\infty }$.