# Why must such a group be dihedral?

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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Is that $U = C_2 \times (C_2 * C_2)$? – Myself Jan 26 '12 at 21:39
Here is rather intuitive guess: it could perhaps be explained by observing that any finite factorgroup of $C_2 * C_2$ is generated by two involutions hence (say) $D_{2n}$ and if $n$ is odd, then I would hope that $D_{4n} = D_{2n}\times C_2$. – Myself Jan 26 '12 at 21:47
@Myself: yes ($y \times (x\ast z)$), though I was thinking of it as a quotient of $C_2 \ast (C_2\times C_2) = x \ast (y\times z)$ – Jack Schmidt Jan 26 '12 at 21:48
Ok, so x,z is dihedral of order 2n, and then if y is not contained in the subgroup generated by x,z, then n must be odd, and we get the result. If y is contained in x,z, then of course the group is x,y,z=x,z so dihedral. Feel free to write it as an answer, and I'll accept. – Jack Schmidt Jan 26 '12 at 21:53
Feels a bit awkward, not having checked that last isomorphism myself, but I've accepted your offer nevertheless :-) – Myself Jan 26 '12 at 21:58

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.