Group completion of $\mathbb{E}_{\infty}$-monoids via tensor products $\newcommand{\K}{\mathrm{K}}$The abelian group completion functor $\K_0\colon\mathsf{CMon}\to\mathsf{Ab}$ satisfies
$$
\K_0(A)
\cong
\mathbb{Z}\otimes_{\mathbb{N}}A,
$$
naturally in $A\in\mathrm{Obj}(\mathsf{CMon})$, where

*

*$\mathbb{Z}$ is the additive monoid of integers (i.e. $\K_0(\mathbb{N})$, the group completion of $\mathbb{N}$);

*$\otimes_\mathbb{N}$ is the tensor product of commutative monoids.


Question. Does the $\mathbb{E}_{\infty}$-group completion functor $\K_0\colon\mathsf{Mon}_{\mathbb{E}_\infty}(\mathcal{S})\to\mathsf{Grp}_{\mathbb{E}_\infty}(\mathcal{S})$ similarly satisfies
$$\K_0(X)\cong QS^{0}\otimes_\mathbb{F}X,$$
where now

*

*$QS^0$, the stabilization of $S^0$, is the $\mathbb{E}_{\infty}$-group completion of $\mathbb{F}=\coprod_{n\in\mathbb{N}}\mathbf{B}\Sigma_{n}$, the groupoid of finite sets and permutations;

*$\otimes_{\mathbb{F}}$ is the tensor product of $\mathbb{E}_{\infty}$-spaces?

 A: Yes, for the same reason. Let me sketch a proof.
1- $QS^0\otimes X$ is group-complete. Indeed, its $\pi_0$ is $\mathbb Z\otimes \pi_0(X)$, and that's a group for the usual reasons. Another way to prove it is to prove that the shear map for $X\otimes Y$ is (the shear map of $X)\otimes Y$, which can be seen by noting that $\otimes$ commutes with coproducts and hence finite products in each variable.
2- There is a natural transformation $X\to QS^0\otimes X$ given by tensoring $\mathbb F\to QS^0$ by $X$, and this induces a natural transformation $X^{gp}\to QS^0\otimes X$.
3- Both sides commute with colimits (a colimit of grouplike $E_\infty$-spaces is grouplike so I don't have to worry about whether I'm talking about colimits in monoids or grouplike monoids), therefore to check that this map is an equivalence, it suffices to do so for $X= \mathbb F$, and for that one it is a tautology.
Another way to phrase this is to use the following sequence of natural equivalences (and using point 1- for the last one):
$X^{gp} = QS^0\otimes_{QS^0} X^{gp} = (QS^0\otimes_\mathbb F X)^{gp}= QS^0\otimes X$
The second natural equivalence comes from the fact that group completion is symmetric monoidal, and $(QS^0)^{gp}\simeq QS^0$.
