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I tend to think dihedral groups are easy to recognize, but I don't quite see why if G is a quotient of $U = \langle x, y, z : x^2 = y^2 = z^2 = 1, yx=xy, zy=yz \rangle$ and G has order 4 mod 8 (so, 4, 12, 20, etc.) then G must in fact be a dihedral group.

This is related to my previous question on coset graphs having 4-cycles, and nearly confirms my suspicion about dihedral groups.

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    Feels a bit awkward, not having checked that last isomorphism myself, but I've accepted your offer nevertheless :-)2012-01-26

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We think of $U$ as $U = y \times (x * z) \cong C_2 \times (C_2 * C_2)$.

In any finite factorgroup $\langle x,z\rangle$ is isomorphic to some dihedral group, say $D_{2n}$. Either $y\in\langle x,z\rangle$ in the factorgroup and we are done, or the quotient is isomorphic to $D_{2n}\times C_2$, which is isomorphic to $D_{4n}$ whenever $n$ is odd.

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    @JackSchmidt I agree! A$n$d I've editted the a$n$swer.2012-01-26