# Quaternions: Difference(s) between $\mathbb{H}$ and $Q_8$

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

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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.

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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.

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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.

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