There is a quaternion group definition which states $$Q_8 = \langle a, b \mid a^4 = 1, a^2 = b^2, ba = a^{-1}b\rangle$$ where $1$ is the identity element. This is suppose to produce eight distinct elements, but I believe I've found a way to do it with four.
Let $a = ([0], [1])$ and $b = ([1], [0])$, where $[0], [1] \in \mathbb{Z}_2$, and let the group operation be addition of the elements of the pairs, $$([w], [x])([y], [z]) = ([w + y], [x + z]).$$ In this case, $1 = ([0], [0]) \neq a \neq b$, $(a^2)^2 = a^4 = 1$, $a^2 = 1 = b^2$, and $ba = ab = a^{-1}b$. Every condition is satisfied.
You could also think of this as two reflections in $D_4$, a square. Two perpendicular reflections $r_1$ and $r_2$ would satisfy all the same properties.
The definition of a quaternion is supposed to produce a more complicated set of elements than this, but I can't figure out why four elements don't cut it. What am I missing about generators or the definition to cause my interpretation to be so different from what was intended?
Because$$\langle a,b\mid a^4=1,a^2=b^2,ba=a^{-1}b\rangle$$is the largest group spanned by two elements $a$ and $b$ such that those relations hold. If your computations are correct, then your group is a quotient of $Q_8$.