Quaternion Group/Group Presentation

Group Presentation of Quaternion Group

The group presentation of the quaternion group is given by:

$\Dic 2 = \gen {a, b: a^4 = e, b^2 = a^2, a b a = b}$

Proof

Let $G = \gen {a, b: a^4 = e, b^2 = a^2, a b a = b}$.

It is to be demonstrated that $\Dic 2$ is isomorphic to $G$.

Consider the Cayley table for $\Dic 2$:

$\begin {array} {r|rrrrrrrr} & e & a & a^2 & a^3 & b & a b & a^2 b & a^3 b \\ \hline e & e & a & a^2 & a^3 & b & a b & a^2 b & a^3 b \\ a & a & a^2 & a^3 & e & a b & a^2 b & a^3 b & b \\ a^2 & a^2 & a^3 & e & a & a^2 b & a^3 b & b & a b \\ a^3 & a^3 & e & a & a^2 & a^3 b & b & a b & a^2 b \\ b & b & a^3 b & a^2 b & a b & a^2 & a & e & a^3 \\ a b & a b & b & a^3 b & a^2 b & a^3 & a^2 & a & e \\ a^2 b & a^2 b & a b & b & a^3 b & e & a^3 & a^2 & a \\ a^3 b & a^3 b & a^2 b & a b & b & a & e & a^3 & a^2 \end {array}$

We have that:

$a^4 = e$

$b^2 = a^2$

$\paren {a b} a = b$

demonstrating that $\Dic 2$ has the same group presentation as $G$.

Hence the result.

$\blacksquare$

Also presented as

The (group) presentation for the quaternion group can also be expressed in the form:

$\Dic 2 = \gen {a, b: a^4 = e, b^2 = a^2 = \paren {a b}^2}$

Sources

• 1996: John F. Humphreys: A Course in Group Theory ... (previous) ... (next): Chapter $4$: Subgroups: Exercise $4$