Noncommutative subalgebras in quaternions Let us consider quaternions $\textsf{H}$: noncommutative associative algebra with division over reals $\mathbb R$. It is a 4-D continuum. We know that it has a (commutative) 2-dimensional subalgebra of complex numbers $\mathbb C$. The question is whether $\textsf{H}$ contains a 2-dimensional $\textit{noncommutative}$ subalgebra? Of course, existence of zero and unity is implied here. The key properties under question are 2D and noncommutativity. To put it differently, whether exist a noncommutative extension of the 2-parametric family $\mathbb C$ being a subset of $\textsf{H}$? Examples and references welcome. Perhaps one important point should be added here. The sought-for family must contain, like quaternions, a commutative $\mathbb R$-field.
 A: I don't have a quick proof, because I'm lacking the details, although I'm pretty sure they can be filled in, but in general, non-commutativity comes from the ijk elements, the "pure vector" quaternions. And for unit quaternions (we might as well limit ourselves to them), if you have $vw \ne vw$ for $v, w \in \Bbb H_0$, you can rotate $v$ to $i$ and $w$ to lie in the $ij$ plane; at that point, $vw$ has to have a $k$ component. So the algebra that contains $v$ and $w$ isn't $2$-dimensional. 
Actually, maybe that is a proof: 
Suppose you have a 2D subalgebra that's noncommutative. Pick elements $v_0, w_0 $ with $v_0w_0 \ne w_0v_0$. Then neither one is pure-real, since real elements lie in the cetner of the quats. They also span the subalgebra, since they're not multiples of one another (non-commuting guarantees this), hence their span is 2D, hence it's the whole subalgebra. 
Let $v_0 = r + v, w_0 = s + w$. Multiplying out, you find that $vw \ne vw$. 
So now you have two pure-vector quaternions that don't commute. Scale them up to unit length, and you've got two noncommuting vectors in the pure-vector quaternion sphere. 
I claim that for any $q$, $q^{-1}vq$ and $q^{-1}wq$ are again unit pure-vector quaternions that don't commute either. That's easy to check. 
So pick a $q$ with $v'= q^{-1}vq = i$; then $w' = q^{-1}wq$ must have nonzero $j$ or $k$ component, or else it'd commute with $i$. And multiplying $i$ by $ai + bj + ck$ shows that the product either ($a \ne 0$) has a real component (violating the dimension-2 requirement) or that $a = 0$ and the product is $-cj + bk$. For this to lie in the $v'w'$ plane, it must be a linear combination of $i$ and $bj + ck$, i.e., it must be a multiple of $bj + ck$, hence $b = -c$ and $c = b$ or $b = c$ and $c = -b$. (The multiple has to be $1$ because ... well, compute the lengths.) Both possibilities lead to $b = c = 0$, a contradiction. 
A: Let $A$ be an $\mathbb R$ subalgebra of $\mathbb H$. This means that $A$ contains a copy of $\mathbb R$ in the center of $A$, and this copy accounts for $1$ dimension of the $2$ alotted.
Then let $x\in A$ be something that's not in that copy of $\mathbb R$. We have that $A=\{\alpha+\beta x\mid \alpha,\beta\in\mathbb R\}$. But clearly multiplication is commutative between elements of this set.
$(\alpha+\beta x)(\gamma+\delta x)=\alpha\gamma+(\alpha\delta+\beta\gamma)x+\beta\delta x^2$
$(\gamma+\delta x)(\alpha+\beta x)=\gamma\alpha+(\delta\alpha+\gamma\beta)x+\delta\beta x^2$
But these two things are equal because of commutativity of multiplication in $\mathbb R$.
