Is the Quaternion ring infinte?

Question

Of course by the definition of Quaternions, it is a finite set and made of 8 elements namely: $Q_8=\{\pm1, \pm i, \pm j, \pm k\}$ (with the known operations). It also can be seen that it is a division ring, but not a field simply because for example: $i\cdot j =-j\cdot i$. However by Wedderburn's theorem, every finite division ring has to be a field. Therefore it pushes $Q_8$ as a division ring to be infinite (since it is not commutative), while as I said it has only 8 elements. I am sure I am making a silly mistake, but I cannot find it. I would be appreciated if someone could point it out to me.

Answer 1

That set of eight elements is a group, but not a (division) ring. It has only one binary operation, multiplication. You can't do $1+i$. – Gerry Myerson 10 mins ago

You are confusing the quaternion group with the quaternion algebra; the former is a finite group with $8$ elements, and the latter is a $4$-dimensional real division algebra with uncountably many elements. – Qiaochu Yuan 9 mins ago