The homotopy of universal Thom spectrum 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?
 A: 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. 
