4
$\begingroup$

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.

  • 0
    Is that $U = C_2 \times (C_2 * C_2)$?2012-01-26
  • 0
    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 $D_{4n} = D_{2n}\times C_2$.2012-01-26
  • 0
    @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)$2012-01-26
  • 0
    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.2012-01-26
  • 0
    Feels a bit awkward, not having checked that last isomorphism myself, but I've accepted your offer nevertheless :-)2012-01-26

1 Answers 1

5

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.

  • 0
    I think y could be in the subgroup generated by x and z without being the identity (it just has to be in the center, so for instance x and z could generate dihedral of order 4), but we are still done :-)2012-01-26
  • 0
    @JackSchmidt I agree! And I've editted the answer.2012-01-26