Let $G$ be a group. Suppose that the map $G \to G$, $g \mapsto g^r$ is a group homomorphism for every $r \in \mathbb N$. Then $G$ is abelian. Is this also true homotopy-theoretically?


(In the ordinary case, this need only hold for $r=2$, because if $(gh)^2 = g^2h^2$, then canceling a $g$ and an $h$ we find that $hg = gh$, i.e. $G$ is abelian).


More precisely, let $G$ be a group-like $E\_n$-space. For each $r \in \mathbb N$, we may pick $m\_r\in E\_n(r)$ where $E\_n$ is an $E\_n$ operad (it doesn't matter which model we choose, nor which $m\_r$ we choose), and look at the map 


$G \to G$, $g \mapsto m\_r(g,\dots,g)$


**Questions:**


1. Is this map always a map of $E\_{n-1}$-spaces?
2. Supposing the answer to (1) is yes, if this $E\_{n-1}$-map can be made into an $E\_n$ map, does it follow that the $E\_n$ structure on $G$ lifts to an $E\_{n+1}$ structure?


I'm not sure I've chosen the right precisification of this question; I'm open to other interpretations.



If a loopspace admits space-level power operations, is is a higher loopspace?
Ok, here is a counterexample. Let $F$ be the fiber of the map $\Omega(K(\mathbb{F}\_2, 2)\stackrel{i\cdot \mathrm{Sq}^1i}{\to} K(\mathbb{F}\_2,5))$, with $\mathbb{E}\_1$-structure as indicated. Then $F$ does not deloop further since $i\cdot\mathrm{Sq}^1i$ is not a loop map. On the other hand, each of the 'power maps' $K(\mathbb{F}\_2, n) \to K(\mathbb{F}\_2,n)$ is either 0 or 1 (depending on the parity of the power), and either way we can complete this to a commutative square with $i \cdot \mathrm{Sq}^1 i$ and then take loops and fibers to deduce that the power maps $F \to F$ are $\mathbb{E}\_1$-maps.


The same argument does not work to show that $F \times F \to F$ is an $\mathbb{E}\_1$-map precisely because the relevant diagram of EM-spaces no longer commutes.


