## Finite group generated by two elements of order 2 is isomorphic to Dihedral Group

Suppose $G=\langle s,t\rangle$ where both $s$ and $t$ has order 2. Prove that $G$ is isomorphic to $D_{2m}$ for some integer $m$.

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 $\displaystyle G\cong D_{2m}=\langle a,b|a^m=b^2=(ab)^2=1\rangle.$