string bordism group and framed bordism group for $d \leq 6$ and $d \geq 7$

Question

Why do the string bordism group and the framed bordism group coincide the same in dimensions lower than 7 ($d = 0,1,2,3,4,5, 6$)?
Why do the string bordism group and the framed bordism group differ from $d=7$ and so on?

In particular. Note that $\Omega_7^{String}=0$ and $\Omega_7^{fr}={Z}_{240}$?

The framed bordism groups $\Omega_d^{fr}$ of manifolds with a framing of the stable normal bundle (or equivalently the stable tangent bundle) are isomorphic to the stable homotopy groups of spheres $π^s_d$.
BString is the homotopy fibre of the map from BSpin given by half of the first Pontryagin class.

References:

$\Omega_0^{fr}=Z$

$\Omega_1^{fr}=Z_2$

$\Omega_2^{fr}=Z_2$

$\Omega_3^{fr}=Z_{24}$

$\Omega_4^{fr}=0$

$\Omega_5^{fr}=0$

$\Omega_6^{fr}=Z_2$

$\Omega_7^{fr}=Z_{240}$

$\Omega_7^{String}=0$

John Rognes · Accepted Answer · 2023-11-23 16:27:52Z

The map $i : * = B\{e\} \to BString$ over $BF = BGL_1(S)$ is $7$-connected, so induces a $7$-connected map $S = M\{e\} \to MString$ of Thom spectra, by the Thom isomorphism and Hurewicz theorem. At the level of homotopy groups, this is $\Omega^{fr}_* \to \Omega^{String}_*$, which is therefore an isomorphism for $* \le 6$ and surjective for $* = 7$.
You can factor the map $i$ through $S^8$, with $S^8 \to BGL_1(S)$ adjoint to $\sigma : S^7 \to GL_1(S)$, so the Thom spectrum over $S^8$ is $S \cup_\sigma e^8$ (see Adams, J(X) IV, Lemma 10.1). Hence the generator $\sigma \in \pi_7(S) = \Omega^{fr}_7$ maps to zero in $\Omega^{String}_7$.

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