# 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]
6 $b^{-1}a^nb = a^n$, i.e., $b^{-1}zb = z$ $z = a^n = b^2$ [SHOW MORE]
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]
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]