TITLE: Show that a finite group G generated by two elements of order 2 is isomorphic to a dihedral group $D_{2n}$ for some n. (Proof Verification)
QUESTION [5 upvotes]: Show that a finite group G generated by two elements of order 2 is isomorphic to a dihedral group $D_{2n}$ for some n. (Proof Verification)
Proof:
Let G be generated by c, b, where $c^2 = b^2 = 1$. Let $a = cb$ be an element of order, say n. The element a is of finite order since G is finite. G is clearly generated by a, b since $c = cbb = ab$ is generated by a, b. Note that $a^{-1} = bc$ since $bca = bccb = 1$. Therefore $bab = bcbb = bc = a^{-1}$. 
Let $\alpha$ be the rotation of $2\pi/n$ degree and $\beta$ the reflection along the line formed by $v_0$ and the origin in $D_{2n}$.
Let $f: G \to D_{2n}$ be defined by $f(a^kb^l) = \alpha^k\beta^l$ for $0 \leq k \leq n$ and $0 \leq l \leq 1$. 
It is clear that $f$ is homomorphic and injective.
Now, we need to prove that $f$ is surjective. It is sufficient to show that f is surjective by showing that G has $2n$ elements since $f$ is injective.
The group G contains 2 subgroups $H_1 = <a>$ and $H_2 = <b>$ of order n and 2, respectively. If we can show that $H_1 \cap H_2 = 1$, then G must contain $2n$ elements. Clearly $b \notin H_1$ . If $a^i = b$, for $1 \leq i \leq n - 1$  then $a^{i-1} = a^ibc = bbc = c$, which is a contradiction to the fact that a is of order n. We also know that $a \neq b$, else c will be the identity, which is a contradiction. This implies that $a^i \notin H_2$
Hence, the result.

REPLY [3 votes]: We write $G=\langle s,t\rangle$, where we have renamed the two generators as $s$, $t$.
Note that $G=\langle st, t\rangle$ since $(st)t=s$. Since $G$ is finite, $st$ has a finite order, say $m$, so that $(st)^m=1_G$. We also have $[(st)t]^2=s^2=1$.
We claim that there are no other relations, other than $(st)^m=t^2=[(st)t]^2=1$.
Suppose to the contrary $sts=1$. Then $sstss=ss$, i.e.\ $t=1$, a contradiction. Similarly if $ststs=1$, $tsststsst=tsst$ implies $s=1$, a contradiction. Inductively, $(st)^ks\neq 1$ and $(ts)^kt\neq 1$ for any $k\geq 1$.
Thus $$G\cong D_{2m}=\langle a,b|a^m=b^2=(ab)^2=1\rangle.$$