# Dicyclic Group is Non-Abelian Group

## Theorem

The dicyclic group $\Dic n$ is a non-abelian group on two generators.

### Corollary

The quaternion group $Q_4$ is a non-abelian group.

## Proof

The dicyclic group $\Dic n$ is defined as follows:

For even $n$, the dicyclic group $\Dic n$ of order $4 n$ is the group having the presentation:

$\Dic n = \gen {a, b: a^{2 n} = e, b^2 = a^n, b^{-1} a b = a^{-1} }$

First it is to be demonstrated that $\Dic n$ is a group.

First we deduce the following:

$(1): \quad b^4 = e$:

 $\ds b^2$ $=$ $\ds a^n$ Definition of Dicyclic Group $\ds \leadsto \ \$ $\ds \paren {b^2}^2$ $=$ $\ds \paren {a^n}^2$ $\ds \leadsto \ \$ $\ds b^4$ $=$ $\ds a^{2 n}$ $\ds$ $=$ $\ds e$ Definition of Dicyclic Group

$(2): \quad b^2 a^k = a^{k + n} = a^k b^2$:

 $\ds b^2 a^k$ $=$ $\ds b^2 a^{2 n} a^k$ Definition of Dicyclic Group: $a^{2 n} = e$ $\ds$ $=$ $\ds b^2 \paren {b^2 a^n} a^k$ Definition of Dicyclic Group: $b^2 = a^n$ $\ds$ $=$ $\ds b^4 a^{n + k}$ $\ds$ $=$ $\ds a^{k + n}$ Definition of Dicyclic Group: $b^4 = e$ $\ds$ $=$ $\ds a^k a^n b^4$ Definition of Dicyclic Group: $b^4 = e$ $\ds$ $=$ $\ds a^k \paren {a^n b^2} b^2$ $\ds$ $=$ $\ds a^k \paren {a^n a^n} b^2$ Definition of Dicyclic Group: $b^2 = a^n$ $\ds$ $=$ $\ds a^k \paren {a^{2 n} } b^2$ $\ds$ $=$ $\ds a^k b^2$ Definition of Dicyclic Group: $a^{2 n} = e$

$(3): \quad j = \pm 1 \implies b^j a^k = a^{-k} b^j$

 $\text {(4)}: \quad$ $\ds a^k b^{-1}$ $=$ $\ds a^{k - n} a^n b^{-1}$ $\ds$ $=$ $\ds a^{k - n} b^2 b^{-1}$ $\ds$ $=$ $\ds a^{k - n} b$

Thus, every element of $\Dic n$ can be uniquely written as $a^k b^j$, where $0 \le k < 2 n$ and $j \in \set {0, 1}$.

The definition of the group product gives:

$a^k a^m = a^{k + m}$
$a^k a^m b = a^{k + m} b$
$a^k b a^m = a^{k - m} b$
$a^k b a^m x = a^{k - m + n}$

Taking the group axioms in turn:

### Group Axiom $\text G 0$: Closure

Let $x, y \in \Dic n$.

Thus $\Dic n$ is closed.

$\Box$

### Group Axiom $\text G 1$: Associativity

Thus $\Dic n$ is associative.

$\Box$

### Group Axiom $\text G 2$: Existence of Identity Element

Thus $e$ is the identity element of $\Dic n$.

$\Box$

### Group Axiom $\text G 3$: Existence of Inverse Element

We have that $e$ is the identity element of $\Dic n$.

Thus every element $...$ of $\Dic n$ has an inverse $...$.

$\Box$

All the group axioms are thus seen to be fulfilled, and so $...$ is a group.

### Generators

The generator of $\Dic n$ is seen to be $\set {a, b}$.

$\blacksquare$