Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

What is the difference between $\mathbb{H}$ and $Q_8$? Both are called quaternions.

share|improve this question

3 Answers 3

up vote 5 down vote accepted

One is the Hamiltonian Quaternions and has many descriptions, perhaps the most important (for things that immediately interest me) is that it is the )up to equivalence) only non-trivial central simple algebra over $\mathbb{R}$--it is also an object of fundamental importance in geometry.

$Q_8$ is the quaternion group. It is of great importance for the many weird properties it has that cause it to be a counterexample to many simple group theoretic questions--it is a non-abelian group all of whose subgroups are normal.

share|improve this answer

A good analogy here is the difference between the complex numbers $\mathbb{C}$ and the cyclic group with 4 elements, which can be realized as the group $\{1, i, -1, -i\}\subset\mathbb{C}$ with complex multiplication.

$\mathbb{C}$ is a 2-dimensional vector space over $\mathbb{R}$, and there is a group inside it denoted $\mathbb{Z}/4\mathbb{Z}=\{\pm 1, \pm i\}$ with complex multiplication. This is the cyclic group with four elements.

Analogously, $\mathbb{H}$ is a 4-dimensional vector space over $\mathbb{R}$, and there is a group called $Q_8$ inside of it, namely the standard basis (with negatives) $\{\pm 1, \pm i, \pm j, \pm k\}\subset\mathbb{H}$ where the operation is quaternion multiplication.

tl;dr -- $Q_8$ sits inside $\mathbb{H}$. The first is known as the quaternion group, and the second thing is the quaternions.

share|improve this answer
The symmetry group of the square has $8$ elements, and whether you take $D_4$ to mean the dihedral group with $4$ or $8$ elements (both conventions exist, the latter is the symmetry group of the square) neither of them is a cyclic group, while $\{1, i, -1, -i\}$ is cyclic. –  Marc van Leeuwen Mar 1 '13 at 9:51
@MarcvanLeeuwen Oops, that's what happens when you write answers at 2am. Thanks for the correction. –  Daenerys Naharis Mar 1 '13 at 19:30

The first is a division ring (also called skew field, which is obviously infinite, and necessarily so by Wedderburn's theorem) and the second is a finite non-abelian group (in fact a subgroup of the multiplicative group of the first). So they are not even the same kind of algebraic structure. Indeed $Q_8$ has only $8$ elements, those unit quaternions that have only one nonzero component. The reason it is called quaternion group is probably that it captures the essence of the definition of multiplication in $\Bbb H$ (the general case follows by $\Bbb R$-bilinearity), and that the quaternions are in fact the most easily descibed context in which one comes across $Q_8$ (but may of course find $Q_8$ in some settings totally unrelated to the quaternions).

Incidentally $Q_8$ shows that the theorem saying that finite subgroups of multiplicative groups of fields (and more generally of integral domains) are cyclic fails for skew fields.

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.