# Why quaternions is a group? [duplicate]

This question already has an answer here:

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.

## marked as duplicate by user1729, ArsenBerk, egreg, Mark, George LawJan 25 '18 at 18:25

• What is it that you do not understand about the answer on the linked question? Your question is identical (just phrased slightly differently)! The answer suggests 3 approaches: Brute force (check all possible products of length 3); simplified brute force (use the automorphism pointed out); or find a matrix representation. Which approach are you having trouble with? You would be better thinking longer about this answer than just asking this question again. – user1729 Jan 25 '18 at 8:54
• @user1729, thanks for link but can you explain for example the idea with automorphism? The author writes "$i\to j \to k$ is automorphism." I know the notion of automorphism but do not know how to use it and what follows from automorphism? – ZFR Jan 25 '18 at 9:01
• You should edit your question to ask this explicitly (or ask new question asking just this, as the current question already has answers). – user1729 Jan 25 '18 at 9:03
• @user1729, Done – ZFR Jan 25 '18 at 9:06
• @user1729, will you explain? – ZFR Jan 25 '18 at 9:15

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.
• $\mathbb{C}^2=\{(z_1,z_2): z_i \in \mathbb{C}\}$, right? What does mean linear mape in this case? – ZFR Jan 25 '18 at 8:56
• @RFZ Yes, that's what $\mathbb{C}^2$. The map $f$ being linear means that if $\alpha\beta\in\mathbb{C}$ and $u,v\in\mathbb{C}^2$, then$$f(\alpha u+\beta v)=\alpha f(u)+\beta f(v),$$but that's not important here. What matters is that I have defined eight maps and that the composition of maps is associative. – José Carlos Santos Jan 25 '18 at 9:06