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] $cbcb^{-1}=(ab)b(ab)b^{-1}=ab^{2}a=ab^{-1}b^{3}a=ab^{-1}az$ $=ab^{-1}a^{-1}a^{p}a^{2-p}z=a(ab)^{-1}a^{p}a^{2-p}z$ $=ac^{-1}za^{2-p}z=aca^{2-p}z=zaca^{2-p}$.
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	$\!a_{1}b_{1}=c_{1}$.			Steps (4), (5)	$\!a_{1}b_{1}=aca^{2-p}c=c^{-1}(caca^{2-p}c)=c^{-1}bcb^{-1}c=c_{1}$
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 $a_{1}=aca^{-2}=a^{2}ba^{-2}$ 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 $a_{1}^{f(p)}=b_{1}^{2}=c_{1}^{2}=a_{1}b_{1}c_{1}=z$, 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	$\langle a_{1},b_{1},c_{1}\rangle$ is isomorphic to a quotient of the dicyclic group with parameter $f(p)$, because it satisfies all the relations for that group, with $a_{1}^{f(p)}=b_{1}^{2}=c_{1}^{2}=a_{1}b_{1}c_{1}=z$.			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 $p=5$.