Noncommutative group of invertible ideals of a ring Let $R$ be a noetherian domain and let $\mathcal{O}$ be an $R$-algebra that is finitely generated and projective as an $R$-module.  The set of invertible fractional ideals of $\mathcal{O}$ is a group under multiplication.  Is this group always abelian?
 A: No; it is not abelian in general.
Let $R$ be a discrete valuation ring with maximal ideal generated by an element $\pi$, let $k := R / \pi R$ be the residue field of $R$ and let $\mathcal{O}$ be the inverse image in $M_2(R)$ of the scalar matrices in $M_2(k)$. Note that $\mathcal{O}$ is free of rank $4$ as an $R$-module, with basis $\{e_{11} + e_{22}, \pi e_{11}, \pi e_{12}, \pi e_{21}\}$ say. So $\mathcal{O}$ is, in particular, finitely generated and projective as an $R$-module.
For every $g \in GL_2(R)$, you have the invertible fractional ideal $\mathcal{O}g$. Note that $\mathcal{O}g \cdot \mathcal{O} h = \mathcal{O} gh$ since $GL_2(R)$ normalises $\mathcal{O}$ inside $M_2(R)$. Note also that $\mathcal{O} g = \mathcal{O} h$ if and only if $gh^{-1} \in \mathcal{O}^\times$, which is the preimage in $GL_2(R)$ of the group of scalar matrices in $GL_2(k)$. So we obtain an injective group homomorphism from $PGL_2(k) \cong GL_2(R) / \mathcal{O}^\times$ into the group $I(\mathcal{O})$ of invertible fractional ideals of $\mathcal{O}$.
In the positive direction, if $K$ is the field of fractions of a Dedekind domain $R$ and if  $\mathcal{O}$ is a maximal order in some finite dimensional separable $K$-algebra, then $I(\mathcal{O})$ is abelian.
See Theorem 12 and Theorem 13(c) of this paper by Frőhlich, which gives a thorough discussion of the Picard group of a noncommutative ring. 
