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:

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

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] $cbcb^{-1} = (ab)b(ab)b^{-1} = ab^2a = ab^{-1}b^3a = ab^{-1}az$ $= ab^{-1}a^{-1}a^pa^{2-p}z = a(ab)^{-1}a^pa^{2-p}z$ $= ac^{-1}za^{2-p}z = aca^{2-p}z = zaca^{2-p}$.
4	$\! bcb^{-1} = caca^{2-p}$			Step (3)	[SHOW MORE] Multiply both sides of Step (2) by $c^{-1}$: $\! bcb^{-1} = c^{-1}zaca^{2-p} = caca^{2-p}$.
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] $c_1$ is conjugate to $c$ and $c^2 = z$ is central, so $c_1^2 = z$. Also, $b_1 = c$, so $b_1^2 = z$.
8	If $p = 3$, then $a_1$ is conjugate to $c$, and hence $a_1^2 = z$			Step (5)	[SHOW MORE] In this case, $2 - p$ becomes $-1$, so we get $a_1 = aca^{-1}$ is also a conjugate of $c$, and $c^2 = z$ is central, so $a_1^2 = c$.
9	If $p = 4$, then $a_1$ is conjugate to $b$, and hence $a_1^3 = z$			Step (5)	[SHOW MORE] In this case, $2 - p = -2$, so $a_1 = aca^{-2} = a^2ba^{-2}$ is a conjugate of $b$, and $b^3 = z$ is central, so $a_1^3 = z$.
10	If $p = 5$ case, then $a_1$ is conjugate to $a$, and hence $a_1^5 = z$			Step (5)	[SHOW MORE] In this case, $2 - p = 3$. $a_1 = aca^{-3} = a^3(a^{-1}b)a^{-3}$ is conjugate to $a^{-1}b$. Since $b^3 = c^2 = abab$, we get $a^{-1}b = bab^{-1}$. Thus, $a_1$ is conjugate to $a$. Since $a^5 = z$ is central, we get $a_1^5 = z$ as well.
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] Steps (6) and (7) gives $b_1^2 = c_1^2 = a_1b_1c_1$, and all equal $z$. Steps (8)-(10) show that this also equals $a_1^{f(p)}$.
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] We only need to eliminate the case that the order of $z$ is exactly one, which is done by exhibiting the groups: SL(2,3) for $p = 3$, binary octahedral group for $p = 4$, and SL(2,5) for $p = 5$.