# 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:

This needs considerable tedious hard slog to complete it.In particular: In my naivety with the below, I am at the moment unaware of the concept of a group presentation being a concept defined from a free group, with the assumption that being a quotient group, a lot of the properties (e.g. associativity, identity) are inherited from the Group Presentation itself. But I need to understand this. So far the only help I've been given is that this area is all "trivial" and "obvious".To discuss this page in more detail, feel free to use the talk page.When this work has been completed, you may remove this instance of `{{Finish}}` from the code.If you would welcome a second opinion as to whether your work is correct, add a call to `{{Proofread}}` the page. |

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

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

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Thus $\Dic n$ is closed.

$\Box$

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

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Thus $\Dic n$ is associative.

$\Box$

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

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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 $\Dic n$.

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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 $\set {a, b}$.

$\blacksquare$

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