Are $i,j,k$ commutative? I am trying to understand quaternions. I read that Hamilton came up with the great equation:
A) $i^2 = j^2 = k^2 = ijk = −1$
In this equation I understand that $i,j,k$ are complex numbers.
Later on, I read that 
B) $ij=k$
C) $ji=-k$
So, if $i,j,k$ are complex numbers, and complex number multiplication is commutative, why are these two equations different? I do understand that quaternion multiplication is non-commutative, but I do not understand why multiplying these complex components are also non-commutative.
Could someone please help me understand what is going on here. I (obviously) am not an expert in mathematics, so a simple explanation would be greatly appreciated.

Suppose $z,z'$ are quaternions. We may write these as $z = a + w$, $z' = a' + w'$ where $a,a'$ are real and $w, w'$ are "purely quarternion" in the sense that if we write out $w$ it is simply some linear combination of $i,j,k$ with no real component. If we identify $i,j,k$ with standard unit vectors in $\mathbb{R}^3$ thus thinking of $w,w'$ as vectors then we have $ww' = -(w \cdot w') + w \times w'$ (where again the first term is the real component, the second term is quaternion one).
$$z z' = (aa' - w \cdot w') + (aw' + a'w + w \times w')$$
$$z' z = (aa' - w \cdot w') + (aw' + a'w + w' \times w)$$
$$ zz' - z'z = w \times w' - w' \times w = 2(w \times w')$$
So commutativity fails precisely when the imaginary quaternion parts aren't colinear. 
