Group theory -- mathematics of symmetries

[lpetrich]

I'd like to get back to generalized dihedral groups, to recalculate these groups' conjugacy classes.

Definition, for parameters m, n, r, s:

aⁿ = e, b^m = a^s, ba = a^rb

r^m = 1 mod n, s*(r-1) = 0 mod n

Product: (aⁱb^j)(a^pb^q) = a^{i + p*r^j}b^j+q

Inverse: (aⁱb^j)^-1 = a^{-i*r^(-j)}b^-j

Conjugate: (a^pb^q)^-1(aⁱb^j)(a^pb^q) = a^{r^(-q)*(i + (r^j-1)*p)}b^j

So element aⁱb^j has conjugates a^i'b^j (same power of b) with
i' = r^q*i + (r^j-1)*p mod n for q = 0 to (m-1) and p = 0 to (n-1)

I'll have to work out some special cases before I can say much more about this problem.

(ETA: edited out extra r condition as unjustified)

 

[lpetrich]

An interesting feature of the r's in the generalized dihedral group is that under multiplication they form a group. This group is a subgroup of the  Multiplicative group of integers modulo n for a positive integer n, group Z(n,*).

That group's elements are all positive integers relatively prime to n, meaning that its order is phi, Euler's phi function. It has a somewhat complicated structure.

For n having prime factors p with power k for each one, product of p^k, the group decomposes as follows:

Z(product of p^k,*) = product of Z(p^k,*)

For odd p, Z(p^k,*) = Z(phi(p^k)) = Z(p^k-1*(p-1))

For p = 2, Z(2,*) = identity group, Z(4,*) = Z(2), and Z(2^k) = Z(2) * Z(2^k-2)

The r's in the generalized dihedral group are in the subgroup of Z(n,*) where every element's order is at most m and may be less than that. The identity element's order is, of course, 1.