# Center of binary von Dyck group has order two

## 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:

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

## 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]