# In the Quaternion group, why does $-1*k = -k$, or any other element for that matter?

The Quaternion group is given as

$$Q_8 = \{ \pm 1, \pm i, \pm j, \pm k \}$$

with $(-1)^2 =1$,$i^2 = j^2 = k^2 = ijk = -1$. Is it by definition, then, that we have $-1*k = -k$, or the same multiplication for any other element for that matter?

• Have you tried making a Cayley table? – Cameron Buie Sep 30 '17 at 22:34

Yes, that is what $-k$ means, by definition. We have the element $-1$ and the element $k$, they commute and we call their product $-k$. It is called that partly because of what it represents in the ring of quaternions (where you can prove that $(-1)\cdot k$ fulfills the defining property of $-k$, the additive inverse of $k$), and partly for brevity.