Are groups algebras over an operad? 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.
 A: No, there is no operad whose algebras are groups. Since there are many variants of operads a more precise answer is that if one considers what are known as (either symmetric or non-symmetric) coloured operads then there is no operad $P$ such that morphisms $P\to \bf Set$, e.g., $P$-algebras, correspond to groups. 
In general, structures that can be captured by symmetric operads are those that can be defined by a first order equational theory where the equations do not repeat arguments (for non-symmetric operads one needs to further demand that the order in which arguments appear on each side of an equation is the same). The common presentation of the theory of monoids is such and indeed there is a corresponding operad. The common presentation of the theory of groups is not of this form (because of the axiom for existence of inverses). This however does not prove that no operad can describe groups since it does not show that no other (super clever) presentation of groups can exist which is of the desired form. 
It can be shown that the category of algebras in $Set$ for an operad $P$ has certain properties that are not present in the category $Grp$ of groups. This does establish that no operad exists that describes groups. 
A: As far as I know, the answer is "no".  The point is that the axioms of an operad must contain no repeated variables (think the associativity or commutativity law, which are written $(xy)z = x(yz)$ and $xy = yz$, or the Jacobi identity $[[x,y],z] + [[y,z],x] + [[z,x],y] = 0.$  
On the hand, the axioms for a group include the axiom $x x^{-1} = 1$, which involves the same variable $x$ twice.
A: 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.
