Center of binary von Dyck group has order two

Statement

Define the group:

$\mathrm{\Gamma }(p,q,r):=⟨a,b,c\mid a^p=b^q=c^r=abc⟩$.

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	$bc{b}^{-1}{c}^{-1}=zac{a}^{2-p}$				[SHOW MORE] $cbc{b}^{-1}=(ab)b(ab)b^{-1}=a{b}^{2}a=a{b}^{-1}{b}^{3}a=a{b}^{-1}az$ $=a{b}^{-1}{a}^{-1}{a}^p{a}^{2-p}z=a(ab{)}^{-1}{a}^p{a}^{2-p}z$ $=a{c}^{-1}z{a}^{2-p}z=ac{a}^{2-p}z=zac{a}^{2-p}$.
4	$bc{b}^{-1}=cac{a}^{2-p}$			Step (3)	[SHOW MORE] Multiply both sides of Step (2) by $c^{-1}$: $bc{b}^{-1}=c^{-1}zac{a}^{2-p}=cac{a}^{2-p}$.
5	Let $a_1=ac{a}^{2-p},b_1=c,c_1=c^{-1}bc{b}^{-1}c$.
6	$a_1{b}_1=c_1$.			Steps (4), (5)	$a_1{b}_1=ac{a}^{2-p}c=c^{-1}(cac{a}^{2-p}c)=c^{-1}bc{b}^{-1}c=c_1$
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 $a_1=ac{a}^{-1}$ 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 $a_1=ac{a}^{-2}=a^{2}b{a}^{-2}$ 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, $2-p=3$. $a_1=ac{a}^{-3}=a^3(a^{-1}b)a^{-3}$ is conjugate to $a^{-1}b$. Since $b^3=c^2=abab$, we get $a^{-1}b=ba{b}^{-1}$. 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 $a_1^{f(p)}=b_1^2=c_1^2=a_1{b}_1{c}_1=z$, where $f(3)=2,f(4)=3,f(5)=5$			Steps (6)-(10)	[SHOW MORE] Steps (6) and (7) gives $b_1^2=c_1^2=a_1{b}_1{c}_1$, and all equal $z$. Steps (8)-(10) show that this also equals $a_1^{f(p)}$.
12	$⟨a_1,b_1,c_1⟩$ is isomorphic to a quotient of the dicyclic group with parameter $f(p)$, because it satisfies all the relations for that group, with $a_1^{f(p)}=b_1^2=c_1^2=a_1{b}_1{c}_1=z$.			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 $p=5$.