Let $G$ be a connected simple connected compact Lie group, and $\Gamma \subset G$ a finite subgroup. Then (the underlying manifold of) $G$ can be framed by right-invariant vector fields, and this framing descends to the quotient manifold $G/\Gamma$. Let $[G]$ and $[G/\Gamma]$ represent the corresponding classes in $\mathrm{Mfr}$, the Thom spectrum of framed cobordism, or use the same notation for their images in $\mathrm{MString}$ or $\mathrm{TMF}$.

In a prior MathOverflow question, I learned that when $G = \mathrm{SU}(2)$, the following equality holds in $\mathrm{Mfr}$:

$$ [G/\Gamma] = [G] \times \#\{\text{conjugacy classes in }\Gamma\}.$$

I can give a physical argument suggesting that the above equality should hold in $\mathrm{TMF}$ for arbitrary $G$.

**Question:** Does the above equality hold in $\mathrm{Mfr}$ for arbitrary $G$? How about in $\mathrm{MString}$ or $\mathrm{TMF}$?

*Edit:* In the comments, Andre Henriques asks about my "physical argument". I will outline it here. It is premised on a conjectural description of $\mathrm{TMF}$ due to Stolz and Teichner, based suggestions of Witten and Segal: the $n$th space of the TMF spectrum is conjecturally equivalent to the moduli space of $(1+1)$D $\mathcal N=(0,1)$ SQFTs with gravitational anomaly equal to $n$ times the gravitational anomaly of a single chiral Majoron. Warning: there as yet is no mathematical definition of "$(1+1)$D $\mathcal N=(0,1)$ SQFT", let alone of their moduli space. But if there were such a mathematical definition, then it would be true that these moduli spaces would compile into an $E_\infty$ ring spectrum, which I will call $\mathrm{SQFT}$, equipped with maps $\mathrm{MString} \to \mathrm{SQFT}$ and $\mathrm{SQFT} \to \mathrm{MF}$ and $\mathrm{SQFT} \to \mathrm{KO}[[q]]$; these maps are among the evidence for the conjecture.

In any case, the String orientation $\mathrm{MString} \to \mathrm{SQFT}$ sends a String manifold $M$ to the $\mathcal{N} = (0,1)$ sigma model with target $M$. (The maps $\mathrm{SQFT} \to \mathrm{KO}[[q]]$ and $\mathrm{SQFT} \to \mathrm{MF}$ correspond respectively to compactifying on $S^1$ and on an elliptic curve, always with nonbounding Spin structure.) The field content of this sigma model consists of a boson (worldsheet scalar) $\phi$ valued in $M$ together with an antichiral fermion (worldsheet spinor) $\psi$ valued in $\phi^*\mathrm{T}_M$. The classical EOMs are that $\phi$ is harmonic and $\psi$ is antiholomorphic.

When $M = G$ is a group manifold, you get a "supersymmetric WZW model" at "level $k=0$". By using the right-invariant framing, you can use a change of coordinates to decouple the boson from the fermion. The sigma model is always a dynamical (nonconformal) QFT, with massive degrees of freedom. But RG flow is a homotopy in $\mathrm{SQFT}$, and you can flow the level-$k$ supersymmetric WZW model to the IR, where, after decoupling, it flows to a level-$k$ full WZW times the SCFT of an antichiral fermion valued in $\mathfrak{g} = \mathrm{Lie}(G)$. In the level-0 case, the WZW term is simply the vacuum, and the sigma model with target $G$ flows in the IR to a free-fermion theory.

The sigma model with target $G/\Gamma$ is the result of gauging the $G$-sigma model with by the manifest $\Gamma$ symmetry. Under RG flow, this will flow to a $\Gamma$-symmetry of the free fermion model. Which symmetry? In the UV, the symmetry is manifestly nonanomalous, and so it is nonanomalous in the IR, and the only natural nonanomalous action is the trivial one.

So in the IR, we are gauging the trivial $\Gamma$-action on the 2D SCFT consisting of an antichiral fermion in $\mathfrak{g}$. Since the action is trivial, it decouples, and the result is $($pure 2D $\Gamma$-gauge theory$) \times \overline{\mathrm{Fer}}(\mathfrak{g})$. Pure finite group 2D gauge theory was studied in detail by Frobenius in the late 19th century. Frobenius showed in particular that pure gauge theory in 2D is equivalent to a direct sum of $\#\{$conjugacy classes$\}$ many copies of the vacuum 2D TQFT.

The claim follows in $\mathrm{SQFT}$, and so in $\mathrm{TMF}$ if you believe that $\mathrm{SQFT} = \mathrm{TMF}$. I don't know whether the claim should or should not hold in $\mathrm{MString}$ or $\mathrm{Mfr}$.

10more comments