Center of binary von Dyck group has order two

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Statement

Define the group:

\Gamma(p,q,r) := \langle a,b,c \mid a^p = b^q = c^r = abc \rangle.

Then the element \! z = a^p = b^q = c^r has order two if either of these hold:

  • q = r = 2
  • (p,q,r) = (3,3,2)
  • (p,q,r) = (4,3,2)
  • (p,q,r) = (5,3,2).

Facts used

  1. Group acts as automorphisms by conjugation
  2. Equivalence of presentations of dicyclic group

Proof for q = r = 2

Follows from fact (2).

Proof for the remaining cases

Much of the proof is common between the cases p = 3,4,5. Thus, with the exception of Steps (8)-(10), all other steps are generic to all p.

Step no. Assertion/construction Given data used Previous steps used Facts used
1 z is in the center z commutes with all the generators since it is a power of each of them
2 c= ab c^2 = abc -- Cancel c from both sides
3 \! bcb^{-1}c^{-1} = zaca^{2-p} [SHOW MORE]
4 \! bcb^{-1} = caca^{2-p} Step (3) [SHOW MORE]
5 Let \! a_1 = aca^{2-p}, b_1 = c, c_1 = c^{-1}bcb^{-1}c.
6 \! a_1b_1 = c_1. Steps (4), (5) \! a_1b_1 = aca^{2-p}c = c^{-1}(caca^{2-p}c) = c^{-1}bcb^{-1}c = c_1
7 b_1^2 = c_1^2 = z Step (5) [SHOW MORE]
8 If p = 3, then a_1 is conjugate to c, and hence a_1^2 = z Step (5) [SHOW MORE]
9 If p = 4, then a_1 is conjugate to b, and hence a_1^3 = z Step (5) [SHOW MORE]
10 If p = 5 case, then a_1 is conjugate to a, and hence a_1^5 = z Step (5) [SHOW MORE]
11 If p \in \{ 3,4,5 \}, we have a_1^{f(p)} = b_1^2 = c_1^2 = a_1b_1c_1 = z, where f(3) = 2, f(4) = 3, f(5) = 5 Steps (6)-(10) [SHOW MORE]
12 \langle a_1, b_1, c_1 \rangle is isomorphic to a quotient of the dicyclic group with parameter f(p), because it satisfies all the relations for that group, with a_1^{f(p)} = b_1^2 = c_1^2 = a_1b_1c_1 = z. Step (11)
13 z^2 = e Fact (2) Follows from the previous step and Fact (2).
14 z has order exactly two, i.e., it is not exactly the identity element [SHOW MORE]