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	$b^{-1}a^{n}b=a^{-n}$, 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 $b^{-1}a^{n}b$. The second interpretation follows from $z=a^{n}$.
6	$b^{-1}a^{n}b=a^{n}$, 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 $b^{-1}a^{n}b=a^{n}$
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 $a^{n}=b^{2}=c^{2}$.