Exercise 1.6.24 (If $G = \langle x, y \rangle$, where $|x| = |y| = 2, x\ne y$, then $G \simeq D_{2n}$)

Proof. Let $t=\mathit{xy}$, where $x,y$ are elements of $G$ of order $2$. Then $x^2=y^2=e$, and $t^{-1}=y^{-1}{x}^{-1}=\mathit{yx}$, so

This shows that $\mathit{tx}=x{t}^{-1}$.

Since $t=\mathit{xy}$, we obtain $⟨x,t⟩=⟨x,\mathit{xy}⟩\subset ⟨x,y⟩$ (indeed, $⟨x,\mathit{xy}⟩$ is the intersection of subgroups of $G$ which contains $x$ and $\mathit{xy}$, and $⟨x,y⟩$ is such a subgroup). Conversely $y=x^{-1}t$, thus $⟨x,y⟩=⟨x,x^{-1}t⟩\subset ⟨x,t⟩$. Therefore

Put $n=|t|=|\mathit{xy}|$. Since $G$ is a finite group, $n<\infty$, and since $x\ne y$, $t\ne 1_G$ thus $n\ge 2$. So

The relation $\mathit{xt}{x}^{-1}=t^{-1}$ implies $x{t}^h{x}^{-1}=t^{-h}$ for all $h\in$. Moreover, $t=x{t}^{-1}{x}^{-1}$, thus $t^h=x{t}^{-h}{x}^{-1}$. This shows that

Then $x{t}^h=t^{-h}x$, so

Indeed, if $k$ is even, then $x^k=1_G$ and ${(-1)}^k=1$, so (3) is true, and if $k$ is odd, then $x^k=x$ and ${(-1)}^k=-1$, so (3) is equivalent to (2) in this case.

Using (3), we obtain

Since $G=⟨x,t⟩$, every element $g\in G$ can be written in the form

where the exponents are integers. We will prove by induction on $l$ that any element $g=t^{a_1}{x}^{b_1}\cdots t^{a_{l-1}}{x}^{b_{l-1}}{t}^{a_l}{x}^{b_l}$ can be written in the form $g=t^h{x}^k$ for some integers $h,k\in \mathbb{Z}$.

If $l=0$, then $g=e=t^0{x}^0$, and if $l=1$, then $g=t^{a_1}{x}^{b_1}$. Assume that for some $l\ge 2$ and for all exponents $a_1,b_1,\dots ,a_{l-1},b_{l-1}\in \mathbb{Z}$, we have $t^{a_1}{x}^{b_1}\cdots t^{a_{l-1}}{x}^{b_{l-1}}=t^u{x}^v$ for some integers $u,v$. Then by (4)

and the induction is done. Furthermore, since $r^n=1_G$ and $s^2=1_G$, we have:

We want to prove that $|G|=2n$.

Suppose that $x\in ⟨t⟩$. Then $x=t^i$ for some integer $i$, where $0\le i<n$. Since $x^2=1_G$, we have $t^{2i}=1_G$, where $|t|=n$. Therefore $n\mid 2i$.

• If $n$ is odd, then $g.c.d(n,2)=1$, thus $n\mid i$. Then $x=t^i=1_G$. This is impossible because $|x|=2$.
• If $n$ is even, then $\frac{n}{2}\mid i$, thus $i=k\frac{n}{2}$, where $0<i<n$, so $k=1$ and $x=t^{\frac{n}{2}}$. Then $y=x^{-1}t=t^{1-\frac{n}{2}}=t^{\frac{n}{2}+1}$. But $y^2=1_G$, thus $t^2=1$. This gives $x\in ⟨t⟩=\{1_G,t\}$, but $x=1_G$ is impossible, and $x=t=\mathit{xy}$ gives $y=1_G$, which is impossible

We can conclude that $x\notin ⟨t⟩$.

We suppose now that $|t|=n>2$. The elements $1_G,t,t^2,\dots ,t^{n-1}$ are distinct, and consequently $x,\mathit{tx},t^{2}x,\dots ,t^{n-1}x$ are also distinct (if $t^{i}x=t^{j}x$, then $t^i=t^j$). Moreover $t^{i}x=t^j$ is impossible, otherwise $x=t^{j-i}\in ⟨t⟩$. This shows that $G$ has exactly $2n$ elements, and every element $y\in G$ is uniquely written in the form $y=t^h{x}^k$, where $0\le h<n,0\le k<2$.

We know from Section 1.2 that, as in (1),

With the same proofs, we obtain that every element $x$ of $D_{2n}$ is uniquely written in the form $x=r^h{s}^k$, where $0\le h<n,0\le k<2$, and

These results show that the map

is an isomorphism:

• If $u=r^h{s}^k$ and $v=r^{h^{\prime }}{s}^{k^{\prime }}$ are any elements in $D_n$, then

  $$\begin{array}{llllll}\hfill \phi (u)\phi (v)& =(t^h{x}^k)(t^{h^{\prime }}{x}^{k^{\prime }})& \hfill & & \hfill & \\ \hfill & =t^{h+{(-1)}^k{h}^{\prime }}{x}^{k+k^{\prime }}& \hfill \text{by (4)}& & \hfill & & \hfill \\ \hfill & =\phi (r^{h+{(-1)}^k{h}^{\prime }}{s}^{k+k^{\prime }})& \hfill & & \hfill & \\ \hfill & =\phi (\mathit{uv})& \hfill \text{by (6)},& & \hfill & & \hfill \end{array}$$

  so $\phi$ is a homomorphism.

• If $u=r^h{s}^k\in \ker <!--\; FUNCTION\; APPLICATION\; -->(\phi )(0\le h<n,0\le k<2)$, then $\phi (u)=t^h{x}^k=1_G$.

  • If $k=1$, then $x=t^{-h}\in ⟨t⟩$: this is impossible.
  • If $k=0$, then $t^h=1$, where $0\le h<n$, thus $k=0$ and $u=e$.

  Therefore $\phi$ is injective.

• $\phi :D_{2n}\to G$ is injective, and $|D_{2n}|=|G|=2n$, therefore $\phi$ is surjective.

We have proved

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