proof that quaternion group is a group

Question

My textbook gives the following definition of a quaternion group: $Q = \{\pm 1, \pm i, \pm j, \pm k\}$, where multiplication is defined by: \begin{align*} 1*x &= x & (-1)*x &= -x & (-1)*(-x)&=x \end{align*} for $x = 1,i,j,k$ (NOTE: x does not include -1,-i,-j,-k). Moreover, \begin{align*} i^2= j^2= k^2 = -1, \end{align*} and \begin{align*} ij&=k & jk&=i & ki&=j, \\ ji&=-k & kj&=-i & ki&=-j. \end{align*}

The question it then poses is: prove that this is a group. i.e. prove associativity, that 1 is the unit element, and that every element has an inverse.

I am stuck at trying to prove even the simple statement that $1*-1=-1$. I have no idea how to prove it, after having fruitlessly shuffled formulas for an hour. Am I missing something?

Answer 1

Hint:

first prove associativity, than, from your rules, for $x=1$ you have $-1*(1)=-1$ and: $$ 1*(-1)=1*(-1*(1))=(1*(-1))*1=(-1)*1=-1 $$

Answer 2

There are a couple of ways of doing this.

1) Brute force. Changing notation, write $e_1 = i, e_2 = j, e_3 = k$. Then the rules above become $e_i e_j = - e_j e_i$ for $i\not = j$ and $e_i^2 = -1$. We want to show that $(e_i e_j)e_k = e_i(e_j e_k)$, and we can at least assume without loss of generality that $i\leq j \leq k$. Some further symmetry will reduce the number of cases you have to check, but it still requires some explicit computation.

2) Introducing the Levi-Civita symbol $\epsilon_{ijk}$ and Kronecker delta $\delta_{ij}$, we can rewrite the defining rules for multipication in $Q$ as \begin{align*} e_i e_j = \epsilon_{ijk} e_k - \delta_{ij} \end{align*} (with implicit summation over $k$). Now just write out $e_i (e_j e_k)$ and $(e_i e_j) e_k$, churn through the calculus, and show that they agree. (Proving associativity for products involving the identity $1$ or the central element $-1$ is trivial.) The trick is noting that \begin{align*} \delta_{ij} e_j &= e_i & \epsilon_{ijk} &= -\epsilon_{jik} & \epsilon_{ijk}\epsilon_{pqr} &= \delta_{jq}\delta_{kr} - \delta_{jr}\delta_{kq} \end{align*} (with implicit summation throughout), which allows us to reduce such expressions formally.

This method is probably overkill, but you'll do this sort of thing countless times if you take a quantum mechanics class (or have done it countless times if you have taken a quantum mechanics class), so you might as well do it now.

3) So, where does this group actually come from? There's a quaternion algebra $\mathbb{H}$ defined as algebra (over $\mathbb{R}$ with basis $1, e_1, e_2, e_3$ satisfying the same relation as for $Q$. (It's clear that $Q$ is associative iff $\mathbb{H}$ is, which would be more useful to us if we knew that $\mathbb{H}$ is associative.) The action of $\mathbb{H}$ on itself induces an embedding $\rho:Q \to SU(2)$. Unwinding this map gives an embedding \begin{align*} 1 &\to \begin{pmatrix}1 & 0 \\ 0 & 1\end{pmatrix} & e_1 & \to \begin{pmatrix} i & 0 \\ 0 & -i\end{pmatrix} & e_2 & \to \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix} & e_3 & \to \begin{pmatrix} 0 & i \\ i & 0 \end{pmatrix} \end{align*} with $\rho(-x) = -\rho(x)$ for all $x$. (See also the Pauli matrices, which is just the same computation on the physics side of things.) The point is that associativity is clear, since $SU(2)$ is a group. I'm not going to slog through the computation here, but there's a similar faithful representation $Q \to SL_2(\mathbb{F}_3)$.

In this particular case, I don't think you can get away from doing some sort of computation, but it's good to know that this group $Q$ comes from somewhere.