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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