Why quaternions is a group?

Question

Let's consider the set consisting of $8$ elements $\{\pm1, \pm i,\pm j, \pm k\}$ with the following multiplication table:

We see that $1$ commutes with any of the $\{i,j,k\}$.

For example, we want to consider the product $(-1)\cdot j$, since $i^2=-1$ then we can rewrite it as: $(-1)\cdot j=(i\cdot i)\cdot j=i\cdot(i\cdot j)=i\cdot k=-j$.

In this example and in many others in order to multiply elements we need to use associativity property. How to prove that associativity is true in this set?

In similar topic I have seen approach using automorphism. However, I was not able to comprehend it. Can anyone explain it please?

Would be grateful if somebody can demonstrate some elementary approach. In my opinion it is definitely important to know.

Answer 1

In short, you can check since there are finite elements. Probably an easier method is to identify elements of the group with their matrix representations and inherit the associativity of matrix multiplication from there.

Answer 2

Your table is incomplete, since it does not contain $-1$, $-i$, $-j$ and $-k$. The complete table is:$$\begin{array}{r|r|r|r|r|r|r|r|r|}\cdot&1&-1&i&-i&j&-j&k&-k\\ \hline1&1&-1&i&-i&j&-j&k&-k\\\hline-1&-1&1&-i&i&-j&j&-k&k\\\hline i&i&-i&-1&1&k&-k&-j&j\\\hline-i&-i&i&1&-1&-k&k&j&-j\\\hline j&j&-j&-k&k&-1&1&i&-i\\\hline-j&-j&j&k&-k&1&-1&-i&i\\\hline k&k&-k&j&-j&-i&i&-1&1\\\hline-k&-k&k&-j&j&i&-i&1&-1\\\hline\end{array}$$Now, the question is: is the operation associative? Of course, you can check it case by case. Or you can identify this structure (that is, this set with this binary operation) with another structure in which you already know that the binary operation is associative. You can, for instance, consider the following lincear maps from $\mathbb{C}^2$ into itself: $\pm\operatorname{Id},\pm I,\pm J,\pm K$, with$$I(z,w)=(iz,-iw)\text{, }J(z,w)=(w,-z)\text{, and }K(z,w)=(iw,iz).$$It is easy to check that the composition table of $\{\pm\operatorname{Id},\pm I,\pm J,\pm K\}$ is the same as above. Since the composition of maps is an associative operation, it follows that the table above defines too an associative binary operation.