# 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.

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Now that Zhen Lin added the tag, no need to keep the request for someone to do it in the post! :-) – Asaf Karagila Apr 21 '12 at 0:11
can't argue with that – Jacob Bell Apr 21 '12 at 0:22
Obviously you are not sitting in room 12A. :-) – Asaf Karagila Apr 21 '12 at 0:25
had to look that one up :P – Jacob Bell Apr 21 '12 at 0:27
Two categorical machines you can use to get groups are Lawvere theories (en.wikipedia.org/wiki/Lawvere_theory) and monads (en.wikipedia.org/wiki/Monad_(category_theory)). – Qiaochu Yuan May 9 '12 at 7:29

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.

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Having learned a bit about operads now, I would quite like to know what these properties of algebras over an operad are! – Zhen Lin Mar 11 '13 at 9:58
I'll need to do some digging to find the precise answer. From what I remember categories of operad algebras have rather strong decomposability properties of morphisms (I know that's not saying much). – Ittay Weiss Mar 11 '13 at 19:07

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.

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that's a good point. maybe more flexible objects, like modular operads or something, might remedy this shortcoming. (well, perhaps not modular operads) – Jacob Bell Apr 21 '12 at 0:21
@donkeykong: Wha? There's a very simple way to encode the group axioms categorically, if that is what you are inclined to do: take the category of all finitely-generated free groups. – Zhen Lin Apr 21 '12 at 0:30
@donkykong: Dear Donkeykong, In case you change your mind, googling "clone universal algebra" should bring up plenty of references, and avoid too many bad sci-fi links. And thanks for your kind words. Best wishes, – Matt E Apr 21 '12 at 0:46
@MattE: The opposite of the category of all finitely-generated free groups precisely encodes all the $n$-ary operations expressible in the signature of the theory of groups, modulo the equations of the theory of groups. If we pass to the skeleton $\mathcal{T}$ of this category then we obtain the Lawvere theory of groups. A group object in a category $\mathcal{C}$ with finite products is a finite-product-preserving functor $\mathcal{T} \to \mathcal{C}$. All this was known before operads were invented. – Zhen Lin Apr 21 '12 at 1:23
@ZL: but can this be interpreted operadically? Or is this already an operadic definition? (in the same sense that an operad should be a functor from the category of trees, or something of that sort) – Jacob Bell Apr 21 '12 at 3:22

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.

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Thanks for posting; this was helpful. – David White Apr 10 at 20:10