What are the operations in quaternions as a division ring?

Question

When I studied quaternions in group theory only the product was defined

Now studying rings, my notes say quaternions are a division ring, But this means that we must have 2 operations: sum and product. How are the operations defined then?

Answer 1

Unfortunately, there are a couple of structures that are refered to as “the quaternions” in abstract algebra, which is no doubt the source of your confusion. To paraphrase Hendrik Lenstra, don’t blame me for the poor nomenclature: I did not create this part of the world...

In group theory, “the quaternions” or the “quaternion group of order $8$” usually refers to the group $$Q_8 = \{\pm 1, \pm i, \pm j, \pm k\}$$ with multiplication given by the rules $i^2=j^2=k^2=ijk=-1$ (and the obvious multiplications by $1$ and $-1$).

In ring theory, though, the quaternions (also known as “the real quaternions”, or “the Hamiltonians”, or simply as $\mathbb{H}$) are the objects of the form $$a + bi + cj + dk,\qquad a,b,c,d\in\mathbb{R}$$ with addition given by $$\Bigl( a+bi+cj+dk\Bigr) \oplus \Bigl( r+si+tj+vk\Bigr) = (a+r) + (b+s)i + (c+t)j + (d+v)k,$$ and multiplication given as if they were polynomials using the rules from $Q_8$ to multiply $i$, $j$, and $k$ with each other. Thus $$\begin{align*} (a+bi+cj+dk)\odot (r+si+tj+vk) &= (ar - bs - ct - dv)\\ &\quad + (as + br + cv - dt)i\\ &\quad + (at + cr + ds - bv)j \\ &\quad+ (av + dr + bt - cs)k. \end{align*}$$ You should then verify that this turns $\mathbb{H}$ into a ring, and that if $a^2+b^2+c^2+d^2\neq 0$, then $$(a+bi+cj+dk)^{-1} = \frac{1}{a^2+b^2+c^2+d^2}\Bigl( a-bi-cj-dk\Bigr),$$ which means you actually have a division ring that is not a field.

You can either think of them as formal sums, or you can represent them with suitable matrices (much like complex numbers can be thought of as formal sums, matrices, ordered pairs, etc). There are a few constructions of them. These are actually the original ones, invented/discovered by Hamilton, which led to his bridge vandalism.

Obviously, the additive structure of $\mathbb{H}$ is also a group, though an abelian one that is simply isomorphic to $\mathbb{R}^4$. You can restrict the coefficients to any subfield of $\mathbb{R}$ and still get a division ring.

Answer 2

The quaternion $q=a+bi+cj+dk$ with $a,\,b,\,c,\,d\in\Bbb R$ has conjugate $q^\ast=a-bi-cj-dk$ satisfying $qq^\ast=q^\ast q=a^2+b^2+c^2+d^2$, which is positive unless $q=0$. So we divide as $q_1/q_2=(q_1q_2^\ast)/(q_2q_2^\ast)$ whenever $q_2\ne0$. While $(q_1/q_2)q_2=q_1$, in general$$q_2(q_1/q_2)=(q_2q_1q_2^\ast)/(q_2q_2^\ast)\ne(q_1q_2q_2^\ast)/(q_2q_2^\ast)=q_1.$$However, $(1/q_2)q_2=q_2(1/q_2)=1$, so the quaternions are a division ring but not a field.

In the quaternion group, the above definition of division gives results such as $1/i=-i,\,i/j=-k,\,i/k=j$. Note in particular that all $8$ elements are invertible, but it's not a ring because it isn't closed under addition. Indeed, Wedderburn's little theorem implies finite division rings commute.