Exercise 1.1.27 (Cyclic subgroup of $G$ generated by $x$)

Proof. Let

• Then $H\subseteq G$ by definition, and $1=x^0\in H$, so $H\ne \varnothing$.

• If $g\in H$ and $h\in H$, then $g=x^n$ and $h=x^m$ for some integers $n$ and $m$. Then by Exercise 19,

  $$\mathit{𝑔h}=x^n{x}^m=x^{n+m},\text{where }n+m\in \mathbb{Z},$$

  so $\mathit{𝑔h}\in H$. This shows that $H$ is closed under multiplication.

• Let $g$ be any element of $H$. Then $g=x^n$ for some integer $n\in \mathbb{Z}$. By Exercise 19, $x^n\cdot x^{-n}=x^{n-n}=x^0=1$, and similarly $x^{-n}\cdot x^n=1$, so

  $$x^n\cdot x^{-n}=x^{-n}{x}^n=1.$$

  This shows that the inverse of $x^{-n}$ in $G$ is

  $${(x^n)}^{-1}=x^{-n},\text{where }-n\in \mathbb{Z}.$$

  Therefore $H$ is closed under inverses.

This proves that $H$ is a subgroup of $(G,\star )$. □