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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?

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    $\begingroup$ Could any downvoters please explain why? This is the first time I’ve posted my own answer to my own question after doing research. $\endgroup$ – Chase Ryan Taylor May 12 at 23:53
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    $\begingroup$ 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. $\endgroup$ – Oscar Lanzi May 13 at 2:19
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    $\begingroup$ 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. $\endgroup$ – Somos May 13 at 2:44
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    $\begingroup$ @OscarLanzi I only just saw this and this joke was definitely underrated $\endgroup$ – Chase Ryan Taylor May 25 at 2:07
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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.

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    $\begingroup$ 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. $\endgroup$ – Oscar Lanzi May 13 at 2:13
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    $\begingroup$ $1$ is the multiplicative identity, so $ik = i 1 k = i(-j^2) k$. That $a(-b) = -ab$ follows from $0 = a 0 = a (b + (-b)) = ab + a(-b)$. $\endgroup$ – Robert Israel May 13 at 12:26
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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 $x^2+1,\ y^2+1,\ z^2+1,\ xyz+1$.

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    $\begingroup$ Exercise: show that this ideal is the whole algebra. $\endgroup$ – Robert Israel May 13 at 2:03
  • $\begingroup$ I’m having to read up on quotient rings (I was never able to fully grasp them) but I really look forward to coming back around to fully understand this. I’ll give you a +1 now of course though. $\endgroup$ – Chase Ryan Taylor May 13 at 2:07
  • $\begingroup$ Could you give me the basic rundown of the “quotient” and “ideal”? (At least I know what $\mathbb R[x,y,z]$ is.) $\endgroup$ – Chase Ryan Taylor May 13 at 3:36
  • $\begingroup$ Hmm, Robert is right.. $\endgroup$ – Berci May 13 at 8:35
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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.

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    $\begingroup$ 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? $\endgroup$ – Chase Ryan Taylor May 13 at 3:19
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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 $\boldsymbol{JIK}\neq-\boldsymbol{U}$ $\implies$ $\j\i\k\neq-1$.

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    $\begingroup$ Using the matrix representation doesn't prove your original claim. $\endgroup$ – Berci May 13 at 0:33
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    $\begingroup$ 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. $\endgroup$ – saulspatz May 13 at 0:57
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    $\begingroup$ @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? $\endgroup$ – Chase Ryan Taylor May 13 at 1:34
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    $\begingroup$ That's not the point. Read my comment again. Look at Berci's answer. It proves you are mistaken. $\endgroup$ – saulspatz May 13 at 3:05
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    $\begingroup$ @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 $2\in A^*$) $(xyz+1,x^2+1,y^2+1,z^2+1)=(1)$. $\endgroup$ – Saucy O'Path May 13 at 16:15
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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 $k$ .

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

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  • $\begingroup$ Could you cite which book you pulled this from? $\endgroup$ – Chase Ryan Taylor May 13 at 17:07
  • $\begingroup$ @ChaseRyanTaylor I didn't use a book. $\endgroup$ – Gregory Nisbet May 13 at 17:31
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    $\begingroup$ @ChaseRyanTaylor, looking back over the proof, I realized that we need to assume that $-1$ commutes with everything, and confirmed that here: en.wikipedia.org/wiki/Quaternion_group $\endgroup$ – Gregory Nisbet May 13 at 17:48
  • $\begingroup$ For clarity, I made up this proof, but realized later that I was missing the key detail that $-1$ commutes with everything. The proof is totally invalid without that fact, which comes from the Wikipedia article. $\endgroup$ – Gregory Nisbet 2 days ago

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