Dicyclic Group is Non-Abelian Group

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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".
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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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