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All of quaternions are, from what I understand, defined simply by $$\newcommand{\i}{\mathrm{i}} \newcommand{\j}{\mathrm{j}} \newcommand{\k}{\mathrm{k}} \i^2=\j^2=\k^2=\i\j\k=-1$$ It is known that quaternion multiplication is not commutative. Does this follow directly from the definition? I suspect this might could be proven via contradiction.

Additionally, the quaternion multiplication (Hamilton product) formula is downright disgusting. This makes the proof all the more daunting.

How should I go about deriving the non-commutativity?

gen-ℤ ready to perish
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    Could any downvoters please explain why? This is the first time I’ve posted my own answer to my own question after doing research. – gen-ℤ ready to perish May 12 '19 at 23:53
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    Scientists have recently discovered a universal principle known as the SE Correlation, whereby the probability of a downvote to be constructively explained is inversely proportional to the probability that the downvote is cast in the first place. This observation leads to the Grand Corrollary: Asking people who actually downvoted to explain why = talking to a wall. – Oscar Lanzi May 13 '19 at 02:19
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    This follows directly from the answers to MSE question 3125084 "If we are handed the presentation $\langle i,j,k \mid i^2=j^2=k^2=ijk \rangle$ and nothing more, can we deduce that this is the quaternion group?". In particular, the quaternion group is not commutative. – Somos May 13 '19 at 02:44
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    @OscarLanzi I only just saw this and this joke was definitely underrated – gen-ℤ ready to perish May 25 '19 at 02:07

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Yes, it does (assuming that the algebra is not trivial, i.e. $1 \ne 0$).

$i - jk = - i^2(i-jk) = -i(i^2 - ijk) = -i(-1 - (-1)) = 0$ so $i = jk$.

Similarly $ k - ij = - (k - ij)k^2 = - (k^2 - ijk) k = 0$ so $ij = k$.

Then $ik = - i j^2 k = - (ij)(jk) = -ki$. So either $ik = 0$ or $i$ and $k$ don't commute. But if $ik = 0$, $ijk = -1 \ne 0 = jik $, so $i$ and $j$ can't commute.

Robert Israel
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    It looks to me as if $ik=-ij^2k$ requires assuming commutation with real factors, which is true but in this context needs to be stated as an assumption. Correct me if I am wrong, thanks. – Oscar Lanzi May 13 '19 at 02:13
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    $1$ is the multiplicative identity, so $ik = i 1 k = i(-j^2) k$. That $a(-b) = -ab$ follows from $0 = a 0 = a (b + (-b)) = ab + a(-b)$. – Robert Israel May 13 '19 at 12:26
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Edit: I know this seems to be wrong in part. Really this has just been a learning experience. Nevertheless some parts still have merit. But most importantly I’m leaving it up because we learn just as much from mistakes as we do from perfection.

Here’s a great trick—use the matrix representations of quaternions: $$\newcommand{\i}{\mathrm{i}} \newcommand{\j}{\mathrm{j}} \newcommand{\k}{\mathrm{k}} \boldsymbol I = \pmatrix{\i&0\\0&-\i} \\ \boldsymbol J = \pmatrix{0&1\\-1&0} \\ \boldsymbol K = \pmatrix{0&\i\\\i&0} \\ \boldsymbol U = \pmatrix{1&0\\0&1}$$

Then just multiply. (I used a calculator.) If quaternion multiplication were commutative, then it should be that

$$\begin{align}\i\j\k &= \j\i\k = -1 \\ \boldsymbol{IJK} &= \boldsymbol{JIK} = -\boldsymbol U \\ \pmatrix{\i&0\\0&-\i} \pmatrix{0&1\\-1&0} \pmatrix{0&\i\\\i&0} &= \pmatrix{0&1\\-1&0} \pmatrix{\i&0\\0&-\i} \pmatrix{0&\i\\\i&0} \\ \pmatrix{0&\i\\\i&0} \pmatrix{0&\i\\\i&0} &= \pmatrix{0&\i\\-\i&0} \pmatrix{0&\i\\\i&0} \\ \pmatrix{-1&0\\0&-1} &= \pmatrix{1&0\\0&-1} \end{align}$$

Obviously this is a contradiction. You can also specifically see that in fact $\boldsymbol{JIK}\neq-\boldsymbol{U}$ $\implies$ $\j\i\k\neq-1$.

gen-ℤ ready to perish
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    Using the matrix representation doesn't prove your original claim. – Berci May 13 '19 at 00:33
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    The matrix representation is simply another way of writing down the quaternions. Anti-commutativity is built in. As Berci says, this does nothing to validate your claim, which is in fact false, as Berci's answer demonstrates. – saulspatz May 13 '19 at 00:57
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    @saulspatz If you look at eq. (19) on Wolfram MathWorld, the statement ${\rm ki}=-{\rm ik}$ implies ${\rm ik}\neq{\rm ki}$ which is non-commutativity, right? – gen-ℤ ready to perish May 13 '19 at 01:34
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    That's not the point. Read my comment again. Look at Berci's answer. It proves you are mistaken. – saulspatz May 13 '19 at 03:05
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    @saulspatz Berci's answer doesn't work, because in $\Bbb R[x,y,z]$ (but, for that matter, in all polynomial rings $A[x,y,z]$ such that $2\in A^*$) $(xyz+1,x^2+1,y^2+1,z^2+1)=(1)$. –  May 13 '19 at 16:15
  • @SaucyO'Path Oooh, I didn't even think about that. And even I remember enough algebra to prove it, which is saying something. – saulspatz May 13 '19 at 17:45
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No, it doesn't follow from the given equations:
Consider the commutative algebra $\Bbb R[x,y,z]$ of polynomials of $3$ variables, and take the quotient by the ideal generated by $x^2+1,\ y^2+1,\ z^2+1,\ xyz+1$.

Berci
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One way to show that not all the "nice" properties of multiplication can apply:

Since we have both $ijk=-1$ and $k^2=-1$, $ij=k$. Now render

$(ij)^2=k^2=-1$

but also, if multiplication is to be both commutative and associative

$(ij)^2=(ij)(ij)=i(ji)j=i(ij)j=i^2j^2=(-1)(-1)=+1$

and we are contradicted. To get back on track, we have to give up either commutavitity or associativity, and in the case of quaternions the former choice is made.

Oscar Lanzi
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    What I’m hearing is that you can derive non-commutativity from the $\mathrm i^2\cdots$ definition, but that it’s not inherent to this definition—it’s a choice. Is this a correct impression of what you’ve said? – gen-ℤ ready to perish May 13 '19 at 03:19
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Yes.

The elements $ij$ and $ki$ do not commute with each other, this can be demonstrated using only the associativity of multiplication and the given collection of identities (201). We don't need to assume the existence of multiplicative inverses of $i, j, k$ in order to get a contradiction. In the proofs of the lemmas below we do assume that multiplication by $-1$ is commutative (202).

$$ i^2 = j^2 = k^2 = ijk = -1 \tag{201} $$

$$ (-1) \cdot a = a \cdot (-1) \;\;\;\;\;\;\text{for all $a$} \tag{202}$$

Show that the two values $ij$ and $ki$ are not equal. Technically, this lemma is completely unnecessary since I'm going to show that $ij$ does not commute with $ki$ , but it's a good sanity check.

Lemma #1 : $ij \neq ki $

Assume the negated goal for the purposes of deriving a contradiction.

$$ ij = ki \tag{NG1} $$

Right multiply by $k$ .

$$ ijk = kik \tag{101} $$ $$ -1 = kik \tag{102} $$

Right multiply by $k$ .

$$ -k = kik^2 \tag{103} $$ $$ -k = -ki \tag{104} $$

Left multiply by $k$ :

$$ -k^2 = -k^2i \tag{105} $$ $$ 1 = i \tag{106} $$ $$ \bot \tag{107} $$

Lemma #2 : $(ij)(ki) \neq (ki)(ij) $

$$ (ij)(ki) = (ki)(ij) \tag{NG2} $$ $$ ijki = ki^2j \tag{111} $$ $$ -i = -kj \tag{112} $$

Right multiply by $j$ .

$$ -ij = -kj^2 \tag{113} $$ $$ -ij = +k \tag{114} $$

Right multiply by $k$ .

$$ -ijk = +k^2 \tag{115} $$ $$ +1 = -1 \tag{116} $$ $$ \bot \tag{117} $$

Greg Nisbet
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