Let $S^0\_p$ be the $p$-adic sphere spectrum. Let $GL\_1(S^0\_p)$ be the set of unit componen of $\Omega^{\infty}S^0\_p$. For any map $ X \to BGL\_1(S\_p^0)$ we get a Thom spectrum call it $Mf$. Now consider the identity map on $GL\_1(S^0\_p)$, the Thom spectrum associated to that map is what I call the universal Thom spectra and denote it by $MGL\_1(S^0\_p)$. 


My question is what are the homotopy groups of this Thom spectrum?



The homotopy of universal Thom spectrum
Assume that $R$ is a connective $E\_\infty$ ring spectrum. Typically $GL\_1(R)$ denotes the set of components in $\Omega^\infty R$ which span $GL\_1(\pi\_0 R)=\pi\_0 R^\times$. I would call the unit component $SL\_1(R)$. I will use this notation below. 


In the formalism of Ando-Blumberg-Gepner-Hopkins-Rezk, the Thom spectrum functor takes based spaces over $BGL\_1(R)$ to $R$-module spectra. Since the identity map $BGL\_1(R)\rightarrow BGL\_1(R)$ is terminal in the domain category its Thom spectrum $\mathcal{M}=MGL\_1(R)$ receives a canonical map from any other such Thom spectrum. Note $MGL\_1(R)$ is not always a terminal object in $R$-module spectra. $MSL\_1(R)$ should satisfy a similar property among such Thom spectra which are $H\pi\_0 R$-orientable. 


Using the ABGHR formalism we can identify $$\mathcal{M}=\Sigma^\infty\_+ EGL\_1(S\_p) \wedge^\mathbb{L}\_{\Sigma^\infty\_+ GL\_1(S\_p)} S\_p.$$ This smash product only involves connective spectra, so $$\pi\_0 \mathcal{M} \cong \pi\_0(\Sigma^\infty\_+ EGL\_1(S\_p)) \otimes\_{\pi\_0(\Sigma^\infty\_+ GL\_1(S\_p))} \pi\_0S\_p\cong \mathbb{Z}\otimes\_{\mathbb{Z}[\mathbb{Z}\_p^\times]}\mathbb{Z}\_p.$$ You can see this by examining the $\mathrm{Tor}$ spectral sequence, but there probably is a more elementary argument. As Neil Strickland pointed out, we can identify this torsion product (I would use the beginning of a free resolution of $\mathbb{Z}$ by free $\mathbb{Z}[\mathbb{Z}\_p^\times]$-modules to calculate this): The torsion product is $$\mathbb{Z}\_p/(gx-x)\_{x\in \mathbb{Z}\_p^\times,x\in \mathbb{Z}\_p}.$$ If $p$ is odd this is 0 and if $p$ is 2 then this is $\mathbb{Z}/2$. This implies $MGL\_1(S\_p)\simeq \*$ for $p$ odd.


Similarly $$\pi\_0 MSL\_1(S\_p)\cong \pi\_0 (\Sigma\_+^\infty ESL\_1 S\_p)\otimes\_{\pi\_0 (\Sigma\_+^\infty SL\_1(S\_p))} \pi\_0(S\_p)\cong \mathbb{Z}\otimes\_\mathbb{Z}\mathbb{Z}\_p\cong \mathbb{Z}\_p.$$ If you take the subspace $G$ of $\Omega^\infty S\_p$ spanned by the components $1+p\mathbb{Z}\_p\subset \mathbb{Z}\_p^\times$ (for $p=2$ this is $GL\_1(S\_2)$), then you get a Thom spectrum with $\pi\_0 MG\cong \mathbb{Z}/p.$ When $p$ is $2$ the completion is unnecessary, $\pi\_0 MGL\_1(S)\cong \mathbb{Z}/2$.


As Neil Strickland pointed out, any $E\_2$-ring spectrum, in particular any $E\_\infty$ ring spectrum, $R$ with $\pi\_0 R\cong \mathbb{Z}/p$ is an $H\mathbb{Z}/p$-algebra and hence a wedge of $H\mathbb{Z}/p$-modules. This implies the mod-$p$ homology is a sum of suspensions of copies of the dual Steenrod algebra. The generators of these copies of the dual Steenrod algebra are in 1-1 correspondence with generators of $\pi\_\*R$ as a graded $\mathbb{Z}/p$-vector space.


I would like to thank Neil Strickland for his helpful comments. 


