<s>[INST] Prove that a group generated by two elements of order $2$, $x$ and $y$, is isomorphic to $D_{2n}$, where $n = |xy|.$ I am completely stuck at the question

Let $G$ be a finite group and let $x$ and $y$ be distinct elements of order 2 in $G$ that generate $G$. Prove that $G \cong D_{2n}$, where $n = |xy|.$

I have proved that

Let $x$ and $y$ be elements of order 2 in any group $G$. If $t = xy$ then $tx = xt^{-1}$.

Can I get some hints?
 [/INST] Personally, I like the following presentation for $D_{2n}$:
$$\langle x,y\mid x^n=y^2=(xy)^2=1\rangle$$ Now assume that the following presentation is given:
$$G=\langle x,y\mid x^2=y^2=(xy)^n=1\rangle$$ So we have:
$$
\begin{align*}
G=\langle x,y&\mid x^2=y^2=(xy)^n=1\rangle\\
\cong
\langle x,y,a&\mid x^2=y^2=(xy)^n=1,a=xy\rangle\\
\cong
\langle x,y,a&\mid x^2=y^2=(xy)^n=a^n=1,a=xy\rangle\\
\cong
\langle x,y,a&\mid x^2=y^2=a^n=1,a=xy\rangle\\
\cong
\langle x,y,a&\mid x^2=y^2=a^n=1,a=xy,(ay^{-1})^2=1\rangle\\
\cong
\langle x,y,a&\mid y^2=a^n=1,a=xy,(ay^{-1})^2=1\rangle\\
\cong
\langle y,a&\mid y^2=a^n=1,(ay^{-1})^2=1\rangle\\
\cong
\langle y,a,b&\mid y^2=a^n=1,(ay^{-1})^2=1,b=y^{-1}\rangle\\
\cong
\langle y,a,b&\mid y^2=a^n=1,(ay^{-1})^2=1,b=y^{-1},b^2=1\rangle\\
\cong
\langle y,a,b&\mid a^n=1,(ay^{-1})^2=1,b=y^{-1},b^2=1\rangle\\
\cong
\langle y,a,b&\mid a^n=b^2=(ab)^2=(ay^{-1})^2=1,b=y^{-1}\rangle\\
\cong
\langle y,a,b&\mid a^n=b^2=(ab)^2=1,b=y^{-1}\rangle\\
\cong
\langle a,b&\mid a^n=b^2=(ab)^2=1\rangle
\end{align*}$$
Which is $D_{2n}$.
</s>