Exercise 1.2.17 (Presentation of group $X_{2n}$)

Proof. Recall that

(see p.27), thus $x=x^4$, and $x^3=1$.

Let $z=x^{a_1}{y}^{b_1}\cdots x^{a_m}{y}^{b_m}$ be any element of $X_{2n}$. Using $\mathit{xy}=y{x}^2$, we can transport every $y$ to the left to obtain

Moreover, writing $j=2k+s,i=3q+r,0\le s<2,0\le t<3$, and using $y^2=1,x^3=1$, we obtain

(a)

Suppose that $n\equiv 0(\mathit{mod}3)$, i.e. $n=3k,k\in \mathbb{N}$. Then $$x^n=y^2=1,\mathit{xy}=y{x}^2⟺x^3=y^2=1,\mathit{xy}=y{x}^{-1}.$$

(Indeed $x^n=y^2=1,\mathit{xy}=y{x}^2$ imply $x^3=1$ by the preamble, thus $\mathit{xy}=y{x}^2=y{x}^{-1}$. Conversely, $x^3=y^2=1,\mathit{xy}=y{x}^{-1}$ imply $x^n={(x^3)}^k=1$ and $\mathit{xy}=y{x}^{-1}=y{x}^2$.)

Therefore

$$X_{2n}=⟨x,y\mid x^n=y^2=1,\mathit{xy}=y{x}^2⟩=⟨x,y\mid x^3=y^2=1,\mathit{xy}=y{x}^{-1}⟩\simeq D_6,$$

so $|X_{2n}|=6$.

(b)

Suppose that $\gcd (n,3)=1$. Then there are integers $u,v$ such that $\mathit{nu}+3v=1$, thus $$x=x^{\mathit{nu}+3v}={(x^n)}^u{(x^3)}^v=1.$$

Since $x=1$,

$$X_{2n}=⟨x,y\mid x^n=y^2=1,\mathit{xy}=y{x}^2⟩=⟨y\mid y^2=1⟩\simeq C_2,$$

so $|X_{2n}|=2$.