Exercise 2.1.14 ($\{x \in D_{2n} \mid x^2 = 1\}$ is not a subgroup of $D_{2n}$ ($n \geq 3$))

Proof. Let $H=\{x\in D_{2n}\mid x^2=1\}.$

Let $s,r$ the generators of $D_{2n}$ such that $|s|=2,|r|=n$ and $s^2=r^n=e,\mathit{rs}=s{r}^{-1}$ (where $n\ge 3$).

Then $s\in H$ and ${(\mathit{sr})}^2=s(\mathit{rs})r=\mathit{ss}{r}^{-1}r=e$, so $\mathit{sr}\in H$. But

has order $n\ge 3$, so $s(\mathit{sr})\notin H$. This shows that $H$ is not a subgroup of $D_{2n}$. □