Left- and right-sided principal ideals have same index? One fact about the Lipschitz integers (quaternions of the form $a + bi + cj + dk$ where $a, b, c, d$ are integers) is that the left-sided ideal generated by any element $Q$ has the same index in the additive group as does the right-sided ideal generated by $Q$.
I know this is a fact, but I don't know a simple abstract reason for it. But surely there must be some simple reason I'm overlooking! Thus, I ask at exactly what level of abstraction this sort of thing cleanly holds:


*

*Is it the case that the left-sided and right-sided ideals generated by any element in any ring have the same index in its additive group?

*Failing that, is this the case in any ring with an involution which preserves ring structure except for reversing the order of multiplication (like the conjugation operation on quaternions)?

*Failing that, is this the case in any ring of the form $R[i, j, k \; | \; i^2 = j^2 = k^2 = ijk = -1]$ where $R$ itself is a commutative ring?

*Failing all that... what is the right level of abstraction for this result?

 A: Counterexample to claim one. Let $A$ be the abelian group of sequences of integers, let $R$ be the endomorphism ring of $A$. Let $X$ be the endomorphism that right-shifts a sequence:
$$X(a_0, a_1, \ldots) = (0,a_0, a_1, \ldots).$$
Then $YX = 1$, where $Y$ is the left-shift operator
$$Y(a_0, a_1, \ldots) = (a_1, a_2, \ldots),$$
so the ideal $R \cdot X$ is the whole ring (index $1$). On the other hand, $X \cdot R$ has infinite index; indeed, the image of any endomorphism of the form $XM$ is contained in the set of sequences whose first coordinate is $0$ (so for instance $Y$ is not in $X \cdot R$).
(My guess is the claim is true under some reasonable small-ness assumptions on the ring, e.g. Noetherian-ness, but I don't know much about non-commmutative algebra, so not touching your other questions.)
A: A simple counterexample to (2) is the ring of upper triangular $2\times 2$ matrices over a finite field $\mathbb{F}_q$. The element $\begin{pmatrix}1&0\\0&0\end{pmatrix}$ generates a right ideal of index $q$ but a left ideal of index $q^2$, and there is an involutory anti-isomorphism $$\begin{pmatrix}a&b\\0&c\end{pmatrix}\mapsto\begin{pmatrix}c&b\\0&a\end{pmatrix}.$$
