Equivalence of presentations of dicyclic group

This article gives a proof/explanation of the equivalence of multiple definitions for the term dicyclic group
View a complete list of pages giving proofs of equivalence of definitions

Statement

Consider the group with presentation:

$G := \langle a,b,c \mid a^n = b^2 = c^2 = abc \rangle$

Then:

• The element $z = a^n = b^2 = c^2 = abc$ is in the center of $G$.
• $z^2$ is the identity element of $G$.
• If we set $a = a, x = b^{-1}$, and denote by $e$ the identity element, $G$ satisfies the relations $a^{2n} = e, x^2 = a^n, xax^{-1} = a^{-1}$. Further, the presentation:

$\langle a,x \mid a^{2n} = e, x^2 = a^n, xax^{-1} = a^{-1} \rangle$

also defines $G$.

Facts used

1. Group acts as automorphisms by conjugation

Proof

We use the notation as in the statement above.

Step no.	Assertion/construction	Facts used	Given data used	Previous steps used	Explanation
1	$z$ is central		$z = a^n = b^2 = c^2$		Since every element commutes with its powers, $a,b,c$ all commute with $z$. Since the group is generated by $a,b,c$, the centralizer of $z$ is the whole group, so $z$ is central.
2	$c = ab$, so the group is generated by $a,b$		$c^2 = abc$		We cancel a $c$ from both sides.
3	$b = aba$		$b^2 = c^2$	Step (2)	Combining $b^2 = c^2$ and Step (2), we get $b^2 = abab$, so $b = aba$.
4	$b^{-1}ab = a^{-1}$			Step (3)	Multiply both sides of Step (3) by $b^{-1}$ on the left and $a^{-1}$ on the right, then interchange the two sides.
5	$b^{-1}a^nb = a^{-n}$, i.e., $b^{-1}zb = z^{-1}$	Fact (1)	$z = a^n$	Step (4)	[SHOW MORE] Raise both sides of Step (4) to the $n^{th}$ power and use Fact (1) to simplify $(b^{-1}ab)^n$ to $b^{-1}a^nb$. The second interpretation follows from $z =a^n$.
6	$b^{-1}a^nb = a^n$, i.e., $b^{-1}zb = z$		$z = a^n = b^2$		[SHOW MORE] Since $a^n = b^2$, $b$ must commute with <amth>a^n</math>, so $b^{-1}a^nb = a^n$
7	$z = z^{-1}$ so $z^2$ is the identity			Steps (6),(7)
8	Setting $a = a$, $x = b^{-1}$, we get $a^{2n} = e, x^2 = a^n, xax^{-1} = a^{-1}$		$z = a^n = b^2$	Steps (4), (7)	[SHOW MORE] By Step (7), $z^2 = e$, and since $z = a^n$, we get $a^{2n} = e$. Since $z = b^2$ and it has order two, we also get $z = b^{-2} = x^2$. Thus, we get $a^n = x^2$. The relation $xax^{-1} = a^{-1}$ follows from Step (4), setting $x = b^{-1}$.
9	Conversely, the original relations can be deduced from $a^{2n} = e, x^2 = a^n, xax^{-1} = a^{-1}$ by setting $b = x^{-1}$, $c = ab$, so the two presentations define the same group				[SHOW MORE] Set $z = a^n$. The relation $a^{2n} = e$ gives $z^2 = e$. So $x^2 = x^{-2} = b^2$. Thus, we get $a^n = b^2$. Since $xax^{-1} = a^{-1}$ we get $b^{-1}ab = a^{-1}$ and rearranging (reversing Step (4) to Step (3)) gives $b = aba$. This gives $b^2 = abab$. Set $c = ab$ and get $b^2 = c^2$, so we get $a^n = b^2 = c^2$.