Let $\Bbb S^0(\Bbb H)\cong \mathrm{SU}(2)$ denote the multiplicateive group of unit quaternions.
This group is non-abelian.
Of course, the subgroups generated by as $\def\<{\langle}\def\>{\rangle}\<1,i\>$, $\<1,j\>$ and $\<1,k\>$ are commutative and also isomorphic to $\Bbb S^0(\Bbb C)\cong\mathrm U(1)$ (the multiplicative group of unit complex numbers).
And there are probably other such subgroups conjugate to the ones named above.
And of course, any subgroup of such a conjugate is also abelian.
Question: Are there other abelian subgroups of $\Bbb S^0(\Bbb H)$?
Best Answer
Slup's answer tells us that all abelian subgroups are contained in maximal abelian subgroups which in turn are products of $n$ circles. The number $n$ is the rank of the group, hence also the number of nodes in the dynkin diagram which is $1$ for $SU(2)$ (or more generally $n = N-1$ for $SU(N)$). It remains to see why we have $n= 1$ in this case. I will type a very 'quaternionic' (as opposed to $SU(N)$-ic) answer to this. (The answer will also point out in which circle group the given abelian group lies and why.)
Let $u$ be a unit quaternion that is completely imaginary, i.e. $u \in \mathbb{R}i \oplus \mathbb{R}j \oplus \mathbb{R}k$. The subspace $\mathbb{C}_u := \mathbb{R} \oplus \mathbb{R}u$ is a subalgebra of $\mathbb{H}$ isomorphic to $\mathbb{C}$ via the isomorphism $u \mapsto i$. We even have that there is an exponential map $\exp: \mathbb{R}u \to \mathbb{C}_u$ that maps the imaginary axis in $\mathbb{C}_u$ to the unit circle in $\mathbb{C}_u$ in the standard way: $\exp$ is defined by the same Taylor series as it always is and convergence works just as always since $\mathbb{H}$ is, from a topological perspective, just $\mathbb{R}^4$.
Now this unit circle in $\mathbb{C}_u$, which I will denote $S^1_u$ is one of the maximal abelian subgroups that we were talking about. Concretely I claim:
Claim: let $A$ be any abelian subgroup of $\mathbb{S}(\mathbb{H})$ containing an element $\alpha 1 + \beta u$ for some $\alpha, \beta \in \mathbb{R}$ then $A \subset S^1_u$.
The proof is very simple: pick arbitrary $x$ not in $S^1_u$ and show that it does not commute with $\alpha 1 + \beta u$.
To simplify this: since we are dealing with unit quaternions $x \not\in S^1_u$ is the same as saying $x \not\in \mathbb{C}_u$. We know that $x$ is of the form $\gamma 1 + \delta u + \epsilon v$ with $\gamma, \delta, \epsilon$ in $\mathbb{R}$ and $v$ a unit quaternion that is both purely imaginary AND not in $S^1_u$. To show that $\alpha 1 + \beta u$ does not commute with $x$ it suffices to show that $u$ does not commute with $v$ since everyting inside $\mathbb{C}_u$ does commute with eachother.
Now we just need to remember that for purely imaginary quaternions $u$ and $v$ we have that $\Re(uv) = \Re(vu), \Im(uv) = - \Im(vu)$ and that $\Im(uv) = 0$ if and only if $u$ and $v$ are real scalar multiples of each other. (Here I use notation $\Re(a + bi + cj + dk) = a$ and $\Im(a + bi + cj + dk) = bi + cj + dk$ for the real and imaginary part of a quaternion respectively.)
These facts are so useful in any computation with quaternions that I expect you have seen them before.