# Quaternion Addition forms Abelian Group

## Theorem

Let $\mathbb H$ be the set of quaternions.

Then $\left({\mathbb H, +}\right)$, where $+$ denotes quaternion addition, is an abelian group.

## Proof

Taking the abelian group axioms in turn:

### G0: Closure

Let:

$\mathbf x_1 = a_1 \mathbf 1 + b_1 \mathbf i + c_1 \mathbf j + d_1 \mathbf k$
$\mathbf x_2 = a_2 \mathbf 1 + b_2 \mathbf i + c_2 \mathbf j + d_2 \mathbf k$

be quaternions.

$\mathbf x_1 + \mathbf x_2 = \left({a_1 + a_2}\right) \mathbf 1 + \left({b_1 + b_2}\right) \mathbf i + \left({c_1 + c_2}\right) \mathbf j + \left({d_1 + d_2}\right) \mathbf k$

So as $a_1, a_2, b_1, b_2$ etc. are all elements of $\R$, then so are $a_1 + a_2, b_1 + b_2$ etc.

So $\left({a_1 + a_2}\right) \mathbf 1 + \left({b_1 + b_2}\right) \mathbf i + \left({c_1 + c_2}\right) \mathbf j + \left({d_1 + d_2}\right) \mathbf k$ is a quaternion.

Hence $\left({\mathbb H, +}\right)$ is closed.

$\Box$

### G1: Associativity

From Matrix Form of Quaternion, we can express a quaternion $\mathbf x$ in the form of a matrix:

$\mathbf x = \begin{bmatrix} a + bi & c + di \\ -c + di & a - bi \end{bmatrix}$

We have that Matrix Entrywise Addition is Associative.

It follows that quaternion addition is also associative

$\Box$

### G2: Identity

The identity element of $\left({\mathbb H, +}\right)$ is:

$\mathbf 0 = 0 \mathbf 1 + 0 \mathbf i + 0 \mathbf j + 0 \mathbf k$

as can be seen:

 $\displaystyle \mathbf 0 + \mathbf x$ $=$ $\displaystyle 0 \mathbf 1 + 0 \mathbf i + 0 \mathbf j + 0 \mathbf k + a \mathbf 1 + b \mathbf i + c \mathbf j + d \mathbf k$ $\displaystyle$ $=$ $\displaystyle \left({0 + a}\right) \mathbf 1 + \left({0 + b}\right) \mathbf i + \left({0 + c}\right) \mathbf j + \left({0 + d}\right) \mathbf k$ $\displaystyle$ $=$ $\displaystyle a \mathbf 1 + b \mathbf i + c \mathbf j + d \mathbf k$ $\displaystyle$ $=$ $\displaystyle \mathbf x$ $\displaystyle$ $=$ $\displaystyle \left({a + 0}\right) \mathbf 1 + \left({b+ 0}\right) \mathbf i + \left({c + 0}\right) \mathbf j + \left({d + 0}\right) \mathbf k$ $\displaystyle$ $=$ $\displaystyle a \mathbf 1 + b \mathbf i + c \mathbf j + d \mathbf k + 0 \mathbf 1 + 0 \mathbf i + 0 \mathbf j + 0 \mathbf k$ $\displaystyle$ $=$ $\displaystyle \mathbf x + \mathbf 0$

$\Box$

### G3: Inverses

Let $\mathbf x = a \mathbf 1 + b \mathbf i + c \mathbf j + d \mathbf k$ be a quaternion.

The inverse of $\left({\mathbb H, +}\right)$ is:

$- \mathbf x = -a \mathbf 1 + -b \mathbf i + -c \mathbf j + -d \mathbf k$

as can be seen:

 $\displaystyle \mathbf -x + \mathbf x$ $=$ $\displaystyle -a \mathbf 1 + -b \mathbf i + -c \mathbf j + -d \mathbf k + a \mathbf 1 + b \mathbf i + c \mathbf j + d \mathbf k$ $\displaystyle$ $=$ $\displaystyle \left({-a + a}\right) \mathbf 1 + \left({-b + b}\right) \mathbf i + \left({-c + c}\right) \mathbf j + \left({-d + d}\right) \mathbf k$ $\displaystyle$ $=$ $\displaystyle 0 \mathbf 1 + 0 \mathbf i + 0 \mathbf j + 0 \mathbf k$ $\displaystyle$ $=$ $\displaystyle \mathbf 0$

and similarly for $\mathbf x + \mathbf -x$.

$\Box$

### Commutativity

Commutativity follows from Real Addition is Commutative.

$\Box$

Thus all the abelian group axioms are seen to be fulfilled.

$\blacksquare$