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

Author: mathtuition88

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