Does the non-commutativity of quaternions follow directly from $\rm i^2=j^2=k^2=ijk=-1$? 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?
 A: 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. 
A: 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 $\boldsymbol{JIK}\neq-\boldsymbol{U}$ $\implies$ $\j\i\k\neq-1$.
A: 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 $x^2+1,\ y^2+1,\ z^2+1,\ xyz+1$. 
A: 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.
A: 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 $k$ .
$$ -ijk = +k^2 \tag{115} $$
$$ +1 = -1 \tag{116} $$
$$ \bot \tag{117} $$
