Are j and k on different imaginary planes than i? I'm trying to understand Quaternions. So I understand that a Quaternion is written like $xi+yj+zk+w$. I also understand that $i^2 = j^2 = k^2 = ijk = -1$, and how that can be used to derive equations such as $ij = k$ and $jk = i$. 
One things that confuses me is that $i$ is not equal to $j$ which is not equal to $k$. I can say $i^2 = -1$ and $j^2 = -1$ but I can't say $ij = -1$.
Correct me if I'm misunderstanding something, but why do they seem to have the same product when squared yet all three must be multiplied to equal the product of any one of them squared? Are they supposed to be on different imaginary planes?
 A: One way to view quaternions is using matrices:
$$
1=\begin{bmatrix}
1&0&0&0\\
0&1&0&0\\
0&0&1&0\\
0&0&0&1
\end{bmatrix}
$$
$$
i=\begin{bmatrix}
0&1&0&0\\
-1&0&0&0\\
0&0&0&-1\\
0&0&1&0
\end{bmatrix}
$$
$$
j=\begin{bmatrix}
0&0&1&0\\
0&0&0&1\\
-1&0&0&0\\
0&-1&0&0
\end{bmatrix}
$$
$$
k=\begin{bmatrix}
0&0&0&1\\
0&0&-1&0\\
0&1&0&0\\
-1&0&0&0
\end{bmatrix}
$$
We can view this as $4$ orthogonal vectors which span a $4$ dimensional subspace of $\mathbb{R}^{16}$.
So yes, these can be viewed as $4$ orthogonal basis vectors, $3$ of which are orthogonal to the reals, thus "imaginary".
A: That is correct, $i$, $j$ and $k$ are contained in three separate complex planes contained in the quaternion numbers. 
Just as the complex numbers $\mathbb{C}$ can be thought of as a 2-dimensional vector space over $\mathbb{R}$ with basis $1,i$, the quaternions $\mathbb{H}$ are a 4-dimensional vector space over $\mathbb{R}$ with basis $1,i,j,k$. 
In particular, the plane spanned by $1,i$, the plane spanned by $1,j$, and the plane spanned by $1,k$ are different subplanes, any two of which intersect each other in the real line spanned by $1$. 
So you can indeed think of these three planes as three separate copies of the complex numbers embedded in the quaternion numbers.
A: The operation of quaternion multiplication (product) is a generalization of multiplication of complex numbers, which in turn is a generalization of multiplication of real numbers. It is not "just" a multiplication. Let's look at it in detail.
There is of course a precise definition and theorems describing its properties but in this context it's enough to know that quaternions can be compared to orthogonal unit vectors. Their [vector] product is equal to a rotation, thus $i*j=k$. With each rotation you move into a new orthogonal dimension. After three rotations you land again on a real axis but the vector points to $-1$. So to illustrate it we can write: $k^2=k*k=i*j*k=1*i*j*k=-1$
The same logic applied to the complex $i$ which represented a rotation by 90 degrees in two dimensional space and so $1*i*i=-1$ as a result of two rotations i.e. by 180 degrees.
I hope now it is obvious why $i*j$ not equals $-1$? After two rotations in four dimensional space we have not arrived back to a real axis yet!
This is absolutely equivalent to the explanation with the matrix representation but I find it more intuitive.
Here are some further reference for those concepts
http://physics.info/vector-multiplication
https://en.m.wikipedia.org/wiki/Cayley–Dickson_construction
