Left- and right-sided principal ideals have same index?

Question

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:

1. 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?
2. 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)?
3. 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?
4. Failing all that... what is the right level of abstraction for this result?

Answer 1

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.)

Answer 2

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 $$\begin{pmatrix}a&b\\0&c\end{pmatrix}\mapsto\begin{pmatrix}c&b\\0&a\end{pmatrix}.$$