I am reading "Algebra 1st Edition" by Michael Artin.
The following proposition is Proposition (8.3) on p.221 in this book.
(8.3) Proposition. The elements $x^n,y^2,xyxy$ form a set of defining relations for the dihedral group.
(8.2) $x^n=1,y^2=1,xyxy=1.$
Proof of Proposition (8.3). We showed in Chapter 5 (3.6) that $D_n$ is generated by elements $x,y$ which satisfy (8.2). Therefore there is a surjective map $\phi:F\to D_n$ from the free group on $x,y$ to $D_n$, and $R=\{x^n,y^2,xyxy\}$ is contained in $\ker\phi$. Let $N$ be the smallest normal subgroup of $F$ containing $R$. Then since $\ker\phi$ is a normal subgroup which contains $R,N\subset\ker\phi$. The mapping property of quotients gives us a homomorphism $\overline{\phi}:F/N\to D_n$. If we show that $\overline{\phi}$ is bijective, the proposition will be proved.
Note that since $\phi$ is surjective, $\overline{\phi}$ is too. Also, in $F/N$ the relations $\overline{x}^n=1,\overline{y}^2=1$, and $\overline{x}\overline{y}\overline{x}\overline{y}=1$ hold. Using them, we can put any word in $\overline{x},\overline{y}$ into the form $\overline{x}^i\overline{y}^j$, with $0\leq i\leq n-1$ and $0\leq j\leq 1$. This shows that $F/N$ has at most $2n$ elements. Since $|D_n|=2n$, it follows that $\overline{\phi}$ is bijective, as required.
Is there any proof of $\#(F/N)=2n$ which doesn't use any group other than $F/N$ itself?