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I understand that quaternion multiplication is non-commutative, but what association does it have.

putting into context when we have the statement

when given general numbers, and algebra such as: $A*B*C$ we generally are taught left to right $(A*B)*C$ even though it does not directly matter, but all of the examples that I have found for multiplying quaternions is only the case of multiplying $q_1*q_2$, and no examples of 3, or more.

so because quaternion multiplication is non-commutative, and probably has a implicit associative (might be using this word our of context) property. what is it?

when given $q_1 * q_2 * q_3$ do I treat this as $q_1 * (q_2 * q_3)$, or as $(q_1 * q_2) * q_3$, or does it actually matter as long as the absolute order is maintained?

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    And if you continue your very interesting line of questioning ("what kind of association property does it have?") the answer would be that they are [alternative](http://en.wikipedia.org/wiki/Alternative_algebra), which is a weaker form of associativity. This is:$(xx)y=x(xy)$ and $y(xx)=(yx)x$ for all $x,y$ in the octonions.2012-05-02

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It doesn't matter - quaternions are associative.

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    @Pete I agree. I think that GEdgar was trying to avoid potential confusion on the OP's part by emphasising that it doesn't matter in which order you multiply the quaternion pairs, but the order of the quaternions themselves does matter.2012-05-04
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Let $K$ be a field, and let $A$ be a not-necessarily-associative-or-unital algebra over $K$. For $x,y,z \in A$, we define the associator

$[x,y,z] = (xy)z - x(yz)$.

The associator defines a $K$-linear map $\varphi: A^3 \rightarrow A$. So the following are equivalent:

(i) $A$ is associative.
(ii) The map $\varphi$ is identically zero.
(iii) For any $K$-basis $\{e_i\}_{i \in I}$ of $A$, all associators $[e_i,e_j,e_k]$ of not necessarily distinct triples of basis elements are zero.

Thus, if $A$ has finite dimension $d$ over $K$, showing its associativity is a purely finite calculation involving checking all $d^3$ associators involving elements of a given basis are zero. This calculation is performed for an arbitrary quaternion algebra over a field $K$ (of characteristic not $2$) in $\S 1.6$ of these notes. This is by no means the best way of showing associativity of quaternion algebras -- e.g. it would be better to embed them, possibly after base extension, into a matrix algebra -- but I find it comforting that one can verify associativity of finite-dimensional algebras "while sleeping", so to speak, if one cares to.

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    I never took a any course on proving mathematical properties, so I will have to take your word for it on this, and according to the up-votes I am presuming that it is accurate.2012-05-04