# Should every group be a monoid, or should no group be a monoid?

Question: What is more convenient/useful? Writing mathematics as if every group is a monoid, or as if these two classes are disjoint?

Additional discussion. Define a monoid as follows.

Defn 1. A monoid is a triple $(X,*,e)$ such that $*$ is an associative binary operation on $X$, and $e \in X$, and $e$ has the property that for all $x \in X$ it holds that $x * e = e * x = x$.

From here, there's at least two ways of defining a group.

Defn 2. A group is a monoid $(X,*,e)$ such that for all $x \in X$ there exists $y \in X$ such that $x*y=e$.

Defn 2'. A group is a quadruple $(X,*,e,i)$ such that $(X,*,e)$ is a monoid, and $i$ is a function $X \rightarrow X$, and for all $x \in X$ it holds that $x * i(x) = e$.

Now I understand that minimalism favors Defn 2, while practitioners of universal algebra favor Defn 3. However, this is not my question.

My question is not: which is preferable, Defn 2 or Defn 2'?

Rather, my question is: which is preferable, definitions like Defn 2 such that every group is a monoid, or definitions like Defn 2' such that no group is monoid? So just to clarify, I want to know: what is more convenient/useful? Writing mathematics as if every group is a monoid, or as if these two classes are disjoint?

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Let's say someone proves both definitions ($2$ and $2'$) are equivalent. Would that answer your question? –  Git Gud May 12 '13 at 10:33
@GitGud, not really. I want to know what is more convenient/useful; writing mathematics as if every group is a monoid, or as if these two classes are disjoint. –  goblin May 12 '13 at 10:34
I think your comment above makes the question clearer. –  Git Gud May 12 '13 at 10:35
@GitGud, thanks I will add it to the question. –  goblin May 12 '13 at 10:36
@user18921, I would say that the most convenient/useful is to simply avoid the pedancy implicit in your question :-) The theory of groups and monoids has developed into one of the major constructions of humankind and this disquisition you are making plays absolutely no role in it. (Of course, in the context of universal algebra or others, the question does have some interest, but if that is the context you are interested in you should make it clear) –  Mariano Suárez-Alvarez May 15 '13 at 7:13

In my opinion Definition 2' is the correct definition (and Definition 2 is not a definition, but rather a characterization), because it follows the general principles how algebraic structures are defined in universal algebra (by operations satisfying equations), and therefore it also directly generalizes to other categories, for example topological spaces. A topological group is not just a topological monoid such that every object has an inverse. We also need that the inverse map is a continuous map (it is an interesting fact that for Lie groups this is automatic). So it is useful when the inverse map belongs to the data.

There are several other reasons why this is important: The subgroup generated by a subset $X$ of a group has underlying set $\{a_1^{\pm 1} \cdots \dotsc \cdots a_n^{\pm 1} : a_i \in X\}$. When you view groups as special monoids, you may forget the $\pm 1$ here.

Different categories should always be considered as disjoint. This helps to organize mathematics a lot. Unfortunately, forgetful functors between these categories are usually ignored, treated as if they were identities, but of course they are not. For example, we have the forgetful functor $U : \mathsf{Grp} \to \mathsf{Mon}$. It turns out that it is fully faithful (in many texts group homomorphisms are defined that way, of course again this is not conceptually correct). It has a left adjoint (Grothendieck construction) as well as a right adjoint (group of units). In particular it preserves all limits and colimits. These properties of $U$ show that often (but not always!) there is no harm when you identitfy $G$ with $U(G)$.

By the way, Definition 2' is conceptually not complete yet. You have to assume $x \cdot i(x) = i(x) \cdot x = e$. This becomes important when you study group objects in non-cartesian categories, aka Hopf monoids, for example Hopf algebras. The axiom for the antipode then contains two diagrams.

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Isn't it a little too strong to say that different categories should always be considered as disjoint? I mean, while it certainly is the case that quite often things are identified too hastily, but is there any harm in considering $Ab$ as a subcategory of $Grp$? –  Ittay Weiss May 12 '13 at 11:03
Martin, I don't get the point you make about "The subgroup generated by a subset." Clearly, the subgroup generated by a subset may differ from the submonoid generated by a subset. –  goblin May 12 '13 at 11:04
@Ittay: Yes, $\mathsf{Ab}$ is a subcategory of $\mathsf{Grp}$. But this shouldn't be seen as some subset relation in the sense of set theory. Rather, there should be reason why we view this as a subcategory. That is, we have to remember the forgetful functor $V : \mathsf{Ab} \to \mathsf{Grp}$. It doesn't preserve colimits. It's not a good idea to identify $A$ with $V(A)$ (unfortunately, of course, this is done in almost every text on group theory). E.g. it's always confusing when people consider direct sums of groups which are abelian. Do they have already applied some forgetful functor? –  Martin Brandenburg May 12 '13 at 11:05
@MartinBrandenburg we are on the same page here. It's just that your call to treat all categories as disjoint surprises me a bit. Is it as a sort of remedy to the all too common (bad) identification all over the place? –  Ittay Weiss May 12 '13 at 11:07
Here is another example: Once I asked on mathoverflow if one can give an example of a diagram in the category of schemes which has no colimit. Almost everybody there, among them great mathematicians, read this subconsciously as "... an example of a diagram in the category of schemes such that the colimit of the underlying locally ringed spaces is not a scheme". Of course, this is not the same! It took some time to resolve this misunderstanding. These wrong interpretations can happen when we ignore forgetful functors (even if they are fully faithful). Actually this happens all the time ... –  Martin Brandenburg May 12 '13 at 11:18

Previous answers are very complete. Let me just add a model theoretic point of view. (I did myself ask that kind of questions recently.)

For the model theorist, your question can be stated as : considering $\mathcal L_1$ the language $\{\ast, e\}$ and $\mathcal L_2$ the language $\{\ast, e, i\}$, should a group be defined as a $\mathcal L_1$-structure (satisfying some $\mathcal L_1$-theory $T_1$) or as a $\mathcal L_2$-structure (satisfying some $\mathcal L_2$-theory $T_2$) ?

If your goal is to talk about group, you should prefer $\mathcal L_2$. For example, considering a $\mathcal L_2$-model $\mathfrak M \models T_2$ of base set $M$, the substructure generated by some $A \subseteq M$ is the subgroup $\langle A \rangle$ of $M$. But taking a $\mathcal L_1$-model $\mathfrak N \models T_1$ makes the substructure generated by $A \subseteq N$ only the monoid generated by $A$ in the implicit monoid $N$ and hence the substructure generated by $A$ is not necessarily model of $T_1$. It shows that the language $\mathcal L_1$ is more suitable to talk about monoids than groups.

We can even go a little further. Consider the language $\mathcal L_0 = \{\ast\}$ . You have enough to state a theory $T_0$ of groups : \begin{aligned} \forall x \forall y \forall z, &\ (x \ast y) \ast z = x \ast (y \ast z) \\ \exists e \forall x, &\ x \ast e = x = e \ast x \\ \forall x \exists y, &\ x \ast y = e = y \ast x . \end{aligned} But because of the second statement, the theory is not $\forall\exists$ (i.e. not every sentence of the theory $T_0$ is equivalent to a sentence $\forall x_1 \dots \forall x_n \exists y_1 \dots \exists y_m, \varphi_0(x_1,\dots,x_n,y_1,\dots,y_m)$ with $\varphi_0$ without quantifier). It has for result that a increasing union of groups isn't necessarily a group (here group in the sense a $\mathcal L_0$-model of $T_0$). That is, groups in that sense does not have the simplest induction limit as a group. Actually, what could go wrong is that the union could have no neutral. As you can imagine, the language $\mathcal L_0$ is then more suitable for associative magmas than for monoids or groups.

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Very good answer! –  Martin Brandenburg May 12 '13 at 19:46
Thanks for the point under my answer. I am eager to know more about Sage. Thanks. +1 –  Babak S. May 14 '13 at 10:15

First, a group is a monoid where every element has a two-sided inverse, not just a one-sided inverse.

Second, whether you define a group to be a monoid where every element has a two-sided inverse, or you define a group to be a quadruple $(G,\circ, e, i)$, where $\circ$ is an associative binary operation on $G$, $e$ is a twos-sided identity for the operation, and $i$ is a unary operation assigning to every element a two-sided inverse, is irrelevant to the question whether or not a group is a monoid. The particularities of a definition are of little importance as to what an actual thing is. The reason that every group is a monoid is that there is a fully faithful forgetful functor from $Grp$ to $Mon$ (the category of groups to the category of monoids). That means that there is a way of seeing every group as giving rise to a monoid. Exactly how that is done may depend on the particularities of the given definition of groups and monoids. But, no matter which definitions you'll choose for the concepts of group and monoid, there will result categories $Grp_1$, $Grp_2$, $Mon_1$, and $Mon_2$, with $Grp_1 \cong Grp_2$ and $Mon_1 \cong Mon_2$, and there will be the corresponding forgetful functors.

So, you may choose any particular way to define something. Category theory will then (if that thing you defined gives rise to a category, which most often is the case (but not always)) give you a precise way to saying when two definitions actually define the same thing (strongly the same means the categories will be isomorphic, weakly the same (which is the kind of sameness we care about in maths) means that categories will be equivalent). And then, whether every blool is a gradnal boils down to the existence of a certain nice functor from $Blool$ to $Gradnal$. This is independent of the choice of definition that gave rise to these categories, as long as different categories of blools (resp. gradnals) are equivalent.

In short: a concept lies not in a definition, but rather in the (too large to exist) equivalence class of equivalent categories. In some cases, categories need to be replaced by $\infty$-categories, making things much more difficult and interesting (e.g., this is in a sense modern algebraic topology: spaces are the same as simplicial sets, as long as one considers the relevant $\infty$-categories).

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If you define a mathematical structure, you normally also want to define the "valid" homomorphisms. If you use only universal Horn expressions to define your structure, then there is a canonical definition for homomorphisms that automatically does the right thing. It says that a homomorphism $\pi:A\mapsto B$ has to satisfy "if $R^A(x_1,\ldots,x_n)$ then $R^B(\pi(x_1),\ldots,\pi(x_n))$" for all relations $R$ over the mathematical structure that you can define by atomic expressions in the given language (like $R(x)\Leftrightarrow x=c$, $R(x,y)\Leftrightarrow f(x)=y$ or $R(x,y)\Leftrightarrow x\leq y$).

It's not always possible to use only universal Horn expressions as axioms (for example, a total order has $\forall x \forall y(x\leq y \lor y\leq x)$ as an axiom), but it's a good idea to do it for the mathematical structures where it is possible. There might be exceptions from this rule, but I just wanted to explain the basic motivation to avoid "there exists" statements (if possible) in the axioms of a mathematical structure.

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Is a Horn expression just alternative nomenclature for a Horn clause? –  goblin May 12 '13 at 13:17
@user18921 A clause is "a disjunction of literals". A Horn expression is a conjunction of Horn clauses where some of the variables may be quantified (both universal or existential). A universal Horn expression is a Horn expression where no existential quantification occurs. –  Thomas Klimpel May 12 '13 at 13:35
Hmmm it's not clear to me. Is $\forall x \exists y(x \geq y \wedge y \leq x)$ a Horn expression? –  goblin May 12 '13 at 13:44
@user18921 Yes, $\forall x \exists y(x\geq y \land y\leq x)$ is a Horn expression. However, it is not a universal Horn expression. –  Thomas Klimpel May 12 '13 at 13:59
Okay I think I get it. What do we call a structure that can be defined via universal Horn expressions? And where I can I learn more? Great answer, by the way. –  goblin May 12 '13 at 14:08

With definition 2', every group is still a monoid, for all pratical intents and purposes. Formally one may be a triple and the other a quadruple, but by simply ignoring the last component (i.e. the map of elements to their inverse), you still get a monoid for every group.

If you want to make that relationship more formal, you can say that you have a map $F \::\: (X,*,e,i) \to (X,*,e)$ which takes every group to a monoid, and an associated map $f \::\: X \to X \,:\, x \to x$ which takes elements of the group to elements of the monoid and which is a monoid-isomorphism.

In situations such is this, its customary to then not distinguish between an element (a group) in the domain of $F$ and its image under $F$.

If you want to formalize this even further, I'm guessting that category theory allows you to do that nicely. Others will have to comment on the details, though - my knowledge of category theory is quite limited...

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No, with def 2' groups are not monoids. The forgetful functor you describe is actually there, is not the identity. See also my answer. –  Martin Brandenburg May 12 '13 at 10:54
@MartinBrandenburg Ok, I don't get it. I my purely set-theoretic view of the world, I fail to see a problem with my $F$ and $f$. Is there a way to understand your argument if one is as ignorant when it comes to category theory as I am? –  fgp May 12 '13 at 11:02
@fgp, it's a bit like the cartesian product of sets. It isn't really associative because of those darn parenthesis that must be moved around a bit before you get the bijection. –  Ittay Weiss May 12 '13 at 11:05
@IttayWeiss Hm, I get that. My problem I guess is that category theory seems to make a big deal out of what I perceive to be a small problem. I need to map $(a,(b,c))$ to $((a,b),c)$ - but so what? They only problem I can see is that set-theory doesn't provide you with a universal function which does that - it'd be "too large". But given some specific cartesian product, the existance of the map (as a function in the sense of set-theory) is always guaranteed as far as I can see. –  fgp May 12 '13 at 11:15
@fgp, the issue is not that the function is too large to exist. It exists as a class. The issue is that you need to know that identifying the two via the obvious way is really harmless. What can go wrong? well, what happens if you now consider an element in the cartesian product of 100 sets, with parenthesis placed in some way. Then you consider the same 100 sets and just rearrange the parenthesis. Is it now obvious that all ways to move parenthesis around to obtain the equivalence between elements in the two sets is independent wrt the order in which you move parenthesis? Try to prove it... –  Ittay Weiss May 12 '13 at 11:21

Note that Def. 2' is a standard type of universal algebraic definition, which means that it requires only equational logic.

On the other hand Def. 2 requires both types of quantification, i.e. stronger logic than Def. 2' (namely: predicate logic).

So Def. 2' appears to be more economical, it needs less of expressive power. At the same time it is not cumbersome or too long. In fact, both definitions are of quite the same length.

Therefore, imho, Def. 2' is better.

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