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) uses::Group acts as automorphisms by conjugation
 * 2) uses::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 $$p$$.