TITLE: Are groups algebras over an operad?
QUESTION [6 upvotes]: I'm trying to understand a little bit about operads. I think I understand that monoids are algebras over the associative operad in sets, but can groups be realised as algebras over some operad? In other words, can we require the existence of inverses in the structure of the operad? Similarly one could ask the same question about (skew-)fields.

REPLY [11 votes]: Groups are not algebras for an operad. (Here, by ‘operad’ I mean the non-symmetric monochromatic version, but the same proof goes through for the symmetric coloured version as well.) Indeed, observe that the category of operads has a terminal object, namely the operad for monoids, and so if $\mathcal{P}$ is an operad, then there must be a functor $\textbf{Mon} \to \textbf{Alg}_{\mathcal{P}}$ that commutes with the underlying set functor. Of course, there is no such functor $\textbf{Mon} \to \textbf{Grp}$. 
Indeed, suppose $F : \textbf{Mon} \to \textbf{Grp}$ is a functor that commutes with the underlying set functor. By general nonsense, $F$ must preserve all limits, so $F$ must carry internal monoids in $\textbf{Mon}$ to internal monoids in $\textbf{Grp}$. However, the Eckmann–Hilton argument says that an internal monoid in $\textbf{Mon}$ is the same thing as a commutative monoid, and an internal monoid in $\textbf{Grp}$ is the same thing as an abelian group. So if there were such a functor $F$, every commutative monoid would automatically be an abelian group with the same binary operation and unit. This is clearly absurd: $\mathbb{N}$ is a commutative monoid but not an abelian group.