If a finite group has a generating set containing two elements of order 2, then it is a dihedral group

Let G be a finite group and let x and y be distinct elements of order 2 in G that generate G. Prove that G \cong D_{2n}, where n = |xy|.

Since G is finite, n = |xy| < \infty. By a previous exercise, x and xy satisfy the relations of the usual presentation for D_{2n}; namely, D_{2n} = \langle r,s \ |\ r^n = s^2 = 1, rs = sr^{-1} \rangle. Moreover, G is generated by x and xy; we know G is generated by x and y and we have y = (x^{-1}) \cdot (xy). From our discussion about D_{2n}, then, every element of G can be written uniquely as x^a(xy)^b for some 0 \leq a < 2 and 0 \leq b < n. We can thus define a mapping \varphi : D_{2n} \rightarrow G by \varphi(s^ar^b) = x^a(xy)^b. This mapping is well defined since every element of D_{2n} has a unique representation as s^ar^b for some 0 \leq a < 2 and 0 \leq b < n. Moreover, \varphi is a homomorphism since \varphi(s^a r^b \cdot s^c r^d) = \varphi(s^{a+c} r^{d-b}) = x^{a+c} (xy)^{d-b} x^a (xy)^b \cdot x^c (xy)^d = \varphi(s^ar^b) \cdot \varphi(s^c r^d). \varphi is also injective, due to the uniqueness of representations of elements in D_{2n} and G in terms of r and s adn x and xy, respectively, and is surjective similarly. Thus G \cong D_{2n}.

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