the comments here indicate that $\pi_3^{st}=\Bbb Z/24$ follows from $\pi_3 M\text{Spin}=0$. anybody know how this is so?

Okay here's a harder one that I can't figure out: Is it true that $BBSpin\simeq Spin/SU$?

Or some other well known homogeneous space of Lie groups?

I can show $Spin\simeq B(SO/U)$... and feel like I can get another representation of $BBSpin$ from this, and maybe some Bott periodicity fiddling or something.

pretty sure Spin/SU is right. starting with the wood cofiber sequence Susp^-1 kO --> Susp^-1 kU --> Susp kO, you can 4-truncate to Susp^-1 (kO[4, infty)) --> Susp^-1 (kU[4, infty)) --> Susp(kO[4, infty)). kO[4, infty) loops down to BSpin, so Susp^-1 (...) loops down to Spin. kU[4, infty) loops to BSU, so Susp^-1 (...) loops down to SU. what's left, Susp kO[4, infty), looks down to B(BSpin), a/k/a SU/Spin

:) what i said at the end is what i mean: SU/Spin, not Spin/SU

Spin/SU does appear somewhere else. i think as Susp^6 kO looped down?

if algebraic topology has a hell, it's full of identities in the dual Steenrod algebra that involve the standard generators on the right-hand side and their conjugates on the left

Ick.