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Let $G$ be a finite group that is generated by $\alpha,\beta\in{G}$ of order 2, such that their product isn't of order 2. Show that $G$ is isomorphic to $D_n$ for some n.

Gyt
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  • What is your definition of $D_n$? – lhf Dec 06 '13 at 10:51
  • Perhaps lhf's question stems from the fact that in some cases finite dihedral groups are defined precisely that way: as groups generated by two non-commutative involutions. – DonAntonio Dec 06 '13 at 14:40

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Hint: Let $\gamma = \alpha \cdot \beta$ and $n$ its order. Show that $G$ is generated by $\{\alpha, \gamma\}$. Then figure out that a relation between $\alpha $ and $\gamma $ is precisely the defining relation for the generators of the Dihedral group $D_n$.

Ittay Weiss
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  • Depending on how $D_n$ is defined, this is trivial: http://en.wikipedia.org/wiki/Dihedral_group#Equivalent_definitions. – lhf Dec 06 '13 at 10:59
  • @lhf I assumed OP was given the definition (or at least had seen) of $D_n$ using generators and relations, and then, yes, it's quite straightforward. – Ittay Weiss Dec 06 '13 at 11:04