This is an open-ended question, as is probably obvious from the title. I understand that it may not be appreciated and I will try not to ask too many such questions. But this one has been bothering me for quite some time and I'm not entirely certain that it's completely worthless, which is why I've decided to ask it here.

Why are groups so immensely more important in mathematics and its applications than semigroups are? I know little of group theory, mathematics and its applications, so I cannot understand how much more important they are. But I know they are because I see how many more people are interested in groups than in semigroups.

I wouldn't ask this question if this fact didn't seem a little strange to me. I know that groups are associated with symmetries, or with automorphisms of structures. Cayley's Theorem tells me that every group can be seen as a set (closed with respect to taking compositions and inverses) of automorphisms of a structure with no constants, functions or relations, i.e. a plain set. I know that the automorphisms of a vector space, a module or a group form a group. Obviously, automorphisms are important.

Then we have inverse semigroups. These are less popular but this I can understand. They are associated with partial symmetries, by the Wagner-Preston Theorem. Partial functions do seem much less used than functions.

But then come semigroups. Just like in the two previous cases, there is an "embedding theorem", which says that every semigroup can be embedded in a semigroup of maps from a certain set into itself. And, analogously to the case of groups, the endomorphisms of a vector space, a module or a group form a semigroup.

This seems to say that semigroups are to endomorphisms what groups are to automorphisms. The "conclusion" would be that

$\frac{\mbox{the importance of semigroups}}{\mbox{the importance of endomorphisms}}=\frac{\mbox{the importance of groups}}{\mbox{the importance of automorphisms}}$

Assuming that endomorphisms are about as important as automorphisms, we get that semigroups are about as important as groups. My feeling is that the assumption is correct.

I understand that I must be oversimplifying something at some point. But what am I oversimplifying and where?

I realize that this is possibly a very dumb question, but my confusion is genuine.

Edit: I have realized, after reading your answers and comments, that I made the mistake of using a very vague term without even attempting to define it. The problem seems to be what importance is. Is it popularity or usefulness, or being used a lot, or interesting or being interesting, or something else, or a mixture of many traits? I'm not going to force my understanding of the word now. But if someone still wants to answer this question, perhaps it would be good idea if the answer contained an attempt at clearing this up. (Or maybe not. I'm not sure!)

Edit: A question similar to mine is dealt with here.

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    $\begingroup$ +1 Please don't hesitate to ask questions like this. Because mathematicians do not usually speak on such topics in more formal settings, it is only natural that they arise in forums like this one. Such questions can certainly lead to very valuable contributions. $\endgroup$ Jan 23, 2012 at 1:36
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    $\begingroup$ My immediate reaction is that I hear more about monoids than about semigroups. That would even fit your analogy better, because when you're looking at endomorphisms of some structure, the identity map surely is included. Furthermore, a monoid is just a category with one object, and categories are hugely popular and important. It might be that most of the interesting things there are to say about monoids can be generalized to categories and therefore are. $\endgroup$ Jan 23, 2012 at 1:45
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    $\begingroup$ Dear ymar, This question is somewhat related to this earlier one, and my answer there provides a partial answer to your question. Regards, $\endgroup$
    – Matt E
    Jan 23, 2012 at 4:09
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    $\begingroup$ #MattE Thank you. I have read it and upvoted! $\endgroup$
    – user23211
    Jan 23, 2012 at 12:00
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    $\begingroup$ @user1729 If you consider white-black move-pairs as transformations between board states, then you have a monoid of transformations. Some move pairs can be undone, some are permanent. This reflects the type of irreversibility you get when talking about left ideals in a monoid of transformations. After you make pawn transformations, you are forever "trapped" in a subset of the whole space of transformations (the ideal, which absorbs further operations on the left). If all moves were reversible, then you could freely travel between board states. $\endgroup$
    – rschwieb
    Jan 7, 2016 at 17:55

8 Answers 8


To add a remark related to Jim Belk's answer and the OP's comments on that answer:

In many naturally occurring situations, including some of those where group theory is particularly useful, endomorphisms are automatically automorphisms.

For example, if $E/F$ is a finite extension of fields, any endomorphism of $E$ which is the identity on $F$ is automatically an automorphism of $E$.

As another example, if $C$ is a Riemann surface of genus at least $2$, then any (nonconstant) endomorphism of $C$ is necessarily an automorphism.

Any endomorphism of a Euclidean space which preserves lengths is necessarily an automorphism.

Another point to bear in mind is that the groups that arise in practice in geometry are often Lie groups (i.e. have a compatible topological, even smooth manifold, structure). One can define a more general notion of Lie semigroup, but if your Lie semigroup has an identity (so is a Lie monoid) and the semigroup structure is non-degenerate in some n.h. of the identity, then Lie semigroup will automatically be a Lie group (at least in a n.h. of the identity). A related remark: in the definition of a formal group, there is no need to include an explicit axiom about the existence of inverses.

To make a point related to Qiaochu Yuan's answer: in some contexts semigroups do appear naturally.

For example, studying the rings of endomorphisms of an object is a very common technique in lots of areas of mathematics. (E.g., just to make a connection to my first point, for genus $1$ Riemann surfaces, there can be endomorphisms that aren't automorphisms, but then genus $1$ Riemann surfaces can also be naturally made into abelian groups --- so-called elliptic curves --- and there is a whole theory, the theory of complex multiplication, devoted to studying their endomorphisms rings.)

As another example, any ring of char. $p > 0$ has a Frobenius endomorphism, which is not an automorphism in general; but the semigroup of endomorphisms that it generates is typically an important thing to consider in char. $p$ algebra and geometry. (Of course, this semigroup is just a quotient of $\mathbb N$.)

One thing to bear in mind is what you hope to achieve by considering the group/semigroup of automorphisms/endomorphisms.

A typical advantage of groups is that they admit a surprisingly rigid theory (e.g. semisimple Lie groups can be completely classified; finite simple groups can be completely classified), and so if you discover a group lurking in your particular mathematical context, it might be an already well-known object, or at least there might be a lot of known theory that you can apply to it to obtain greater insight into your particular situation.

Semigroups are much less rigid, and there is often correspondingly less that can be leveraged out of discovering a semigroup lurking in your particular context. But this is not always true; rings are certainly well-studied, and the appearance of a given ring in some context can often be leveraged to much advantage.

A dynamical system involving just one process can be thought of as an action of the semigroup $\mathbb N$. Here there is not that much to be obtained from the general theory of semigroups, but this is a frequently studied context. (Just to give a perhaps non-standard example, the Frobenius endomorphism of a char. $p$ ring is such a dynamical system.) But, in such contexts, precisely because general semigroup theory doesn't help much, the tools used will be different.

E.g. in topology, the Lefschetz fixed point theorem is a typical tool that is used to study an endomorphism of (i.e. discrete dynamical system on) a topological space. Interestingly, the same formula is used to study the action of Frobenius in char. $p$ geometry (see the Weil conjectures). So even in contexts such as action of the semigroup $\mathbb N$, there is some coherent philosophy that can be discerned --- it is just that it is supplied by topology rather than algebra, since the algebra doesn't have all that much to say.

I think the conclusion to be drawn is not to be too doctrinaire, and to be sensitive to the actual mathematical contexts in which and from which the various notions of group, semigroup, automorphism, and endomorphism arise and have arisen.

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    $\begingroup$ This is a great answer, thank you. I can't say I understand every word you've written but it did give me a lot of food for thought. $\endgroup$
    – user23211
    Jan 24, 2012 at 0:05

It is true that groups are much more important in mathematics than semigroups. There are two basic reasons for this:

  1. Groups are the mathematical embodiment the concept of symmetry. In the examples you give, you seem to equate the idea of symmetry with automorphisms of algebraic objects. However, most symmetries that mathematicians care about are geometric or analytic, such as the rigid symmetries of a polyhedron, the deck transformations of a covering space, the automorphisms of a graph, the continuous symmetries of a system of differental equations, the isometries of a metric space, and so forth. In these cases, there isn't an obvious analogue of "endomorphism" that could be used to form a semigroup. Even if there were, the semigroup wouldn't be as useful, because the most interesting endomorphisms are the automorphisms.

  2. Algebraic objects tend to have many more endomorphisms than automorphisms, which makes it much easier to understand the automorphism group than the endomorphism semigroup. Moreover, the definition of a group is "rigid" enough that it leads to a rich structure theory, which makes it much easier to investigate the structure of a given group than the structure of a given semigroup. Indeed, one of the first things you would want to understand for a semigroup would be the structure of the group of units.

Of these two reasons, I would say that the first explains why groups are so much more important to mathematics in general, while the second helps to explain why groups are more important even within the context of abstract algebra.

  • $\begingroup$ Ad 1. I was determined, when I wrote my question, to make sure I'm not saying "automorphisms of algebraic structures". I did use examples from algebra because I know too little about other examples to be sure I'm not talking nonsense. What I did mean by "automorphisms" is something that I believe category theory deals with. I know nothing about category theory so I didn't use the term in my question either. $\endgroup$
    – user23211
    Jan 23, 2012 at 11:46
  • $\begingroup$ And to add to my previous comment, I know that category theory has the concept of endomorphism, so I would guess it should be appliable outside algebra. I can't know exactly where of course because I only have a vague understanding of what categories and morphisms are. $\endgroup$
    – user23211
    Jan 23, 2012 at 11:59
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    $\begingroup$ This seems more like an argument/explanation for the ubiquity or convenience of groups, rather than importance per se. As for part (1), which I agree with to som extent, what about the inverse semigroups that arise from groupies of partial symmetries? $\endgroup$
    – user16299
    Jan 23, 2012 at 16:30
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    $\begingroup$ that should have been "groupoids" - curse this gutter-minded autocorrect ... $\endgroup$
    – user16299
    Jan 23, 2012 at 16:50

I disagree that groups are more important than semigroups. For example, the multiplication on a ring without identity turns it into a semigroup, and rings are incredibly important in mathematics. In fact a ring without identity is nothing more than a semigroup internal to the category of abelian groups with tensor product.

What I do believe to be the case is that groups were the first to be historically studied because it is more natural to think about isomorphisms than endomorphisms (it is not at all obvious from the perspective of our mathematical ancestors that a non-isomorphism is a useful thing to think about) and that groups are easier to study since they have more structure (e.g. the representation theory of finite groups is much easier than that of monoids or semigroups; see this question).

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    $\begingroup$ @ymar: the larger point is that when we study rings we acknowledge that endomorphisms are important; if we like monoids because we like endomorphisms of sets, then we like rings because we like endomorphisms of abelian groups, and there is a Cayley's theorem for rings to this effect. $\endgroup$ Jan 23, 2012 at 1:47
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    $\begingroup$ @ymar: it is not particularly hard to work out. Any ring embeds into the endomorphism ring of its underlying abelian group by left multiplication. $\endgroup$ Jan 23, 2012 at 1:49
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    $\begingroup$ @Mariano: sure. For me "ring" means "ring with unit" by default and I use "rng" to denote "rings without unit" because I think it's cute. $\endgroup$ Jan 23, 2012 at 2:14
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    $\begingroup$ I don't agree that just because "the multiplication on a ring without identity turns it into a semigroup, and rings are incredibly important in mathematics" then that automatically makes semigroups important. Just because semigroups arise in some context that is important doesn't automatically make semigroups (as an independent subject of study) important. It seems sort of similar to saying "the integers form a group under addition, and the integers are incredibly important in mathematics" so that makes groups important. $\endgroup$
    – Ted
    Jan 23, 2012 at 2:48
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    $\begingroup$ @Ted: again, the larger point is not only that you can form a semigroup (in fact a monoid for rings with identity, so I'll just say that) from a ring, but that a ring is the same thing as a monoid internal to abelian groups: there is a two-way implication here. Any monoidal category (en.wikipedia.org/wiki/Monoidal_category) carries an internal notion of monoid, and it's a funny fact about mathematics that historically the most important groups have been the ones in $\text{Set}$ while the most important monoids have been the ones in $\text{Ab}$. $\endgroup$ Jan 23, 2012 at 2:50

I had thought about this for a long time, this is what I got.

I don't think that semigroups are less important than groups, mostly because groups are semigroups too. I think that many people tend to disregard semigroups because we cannot prove lot of staff with semigroups axioms, by the way the same applies to groups axioms: every equation in a group has always a unique solution, group homomorphisms preserve identities and inverses, and few more. Many other theorems can be proven if we add some hypothesis, for instance if we add that a group is also finite we can prove a lot of stuff (Lagrange, Cauchy anb Sylow theorem for instance), other things can be proven if we require that our group has also a topological structure compatible with the group-structure. The point is that to prove significant results in all mathematics we have to add more properties then the axioms of groups (but same holds for other structures as topological spaces). So to produce significant results about semigroups we have to consider semigroups with more structure then those of semigroups: one way is to add properties to the operation (for instance adding identity or inverses), another way is to give an action (or representation) of our semigroup (for example if we made act our monoid on an abelian group (with the same underlying set) we get a rings [with or without identities, depending if our semigroups are monoids too]).

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    $\begingroup$ To build on what you've said... Group theory is a subtheory of the theory of semigroups, so groups can only end more important than semigroups if a subtheory S of a theory T can come as equally important to the theory T. This will only happen, I think, if the complementary subtheory of S, T-S has no importance whatsoever. In this case, the parts of the theory of semigroups which don't concern groups, do have importance, so the complementary subtheory has importance. Thus, one might reason that the theory of semigroups has more importance. $\endgroup$ Feb 4, 2012 at 20:58
  • $\begingroup$ @ymar: For the same reason, time zone, I did not see this question earlier. I agree with what ineff said. I believe there are more studies on groups than semigroups because semigroups lack of inverses. That's why inverse semigroups play a significant role in semigroup theory. I myself pay attention to Lallement's book because it contains applications so I have things to play with. To answer your question, all groups are semigroups. Groups are special cases of semigroups. Group theory is a branch of semigroup theory. $\endgroup$
    – Nobody
    May 1, 2012 at 4:25

I would take a step back and talk about magmas. A magma $M$ is a set of objects and a "product" $*\colon M\times M \rightarrow M$. We can think of a magma as a collection of (binary) trees for some set. In particular, a magma is basically a notion that describes "syntax" or "position relations" for some class of objects. But the magma axioms are very weak. A magma includes all the "position relations", and none of the elements are (necessarily) special in any way. In fact, just about the only thing you can prove about all magmas at once is that $x = x$.

Note that a semigroup is an associative magma. In particular, that means that some trees are to be treated as equal to other trees. Super, but none of the elements are (necessarily) special in anyway.

In the magma $\rightarrow$ semigroup $\rightarrow$ group hierarchy, the group is the first class which imposes special conditions on its objects, and not merely the product. It captures some notion of "semantics" for the elements beyond merely saying that "these two paths go to the same object". (Arguably, the elements themselves are "path-like" objects, and group theory is what we get when these two distinct notions of "path-ness" interact)


There is another structure that is weaker than groups, but is important. Those are quasigroups. Quasigroups are informally "magmas with inverses".

I think it would be unfair to compare two structures, when one has more axioms than the other. That's why we should look more at why quasigroups provide a richer theory than that of semigroups - which they certainly do.

Quasigroups connect to combinatorics, for example, Cayley table of any finite quasigroup is a Latin square, and conversely. They have geometric interpretations. The overall theory I think is weaker than that of groups, it might be a combination of reasons, I think it's because they are looked down upon, are harder to work with, and the results are more technical and more combinatorial. I don't think the theory can rival that of groups, but it's certainly rich enough to be useful, which I don't think semigroup theory is.

The associativity of groups is in my opinion just a missing link, a very convenient simplification.

Concluding, I think the reason why groups give a much better theory than semigroups is the operation of inverse.

  • $\begingroup$ Late to the party I know, but I'm just re-embarking on a tour of abstract algebra at the moment: it's worth adding that quasigroups are not necessarily associative, so while quasigroups sit between magmas and groups the same as semigroups do, and quasigroups and semigroups overlap, their existence is independent of semigroups. $\endgroup$ Nov 3, 2021 at 10:49
  • $\begingroup$ ... and of course there is also the concept of an algebra loop, which is a quasigroup with an identity -- therefore a structure which fulfils all the qualities of a monoid without the criterion of associativity. $\endgroup$ Nov 3, 2021 at 10:51
  • $\begingroup$ @PrimeMover monoids don't necessarily have inverses $\endgroup$
    – Jakobian
    Nov 3, 2021 at 12:40
  • $\begingroup$ Nor do algebra loops. An algebra loop is a quasigroup with an identity. $\endgroup$ Nov 3, 2021 at 12:43

From my point of view, the "most natural" structure is to be found in the middle of the two structures, namely the one of monoid. A monoid is just a set with an associative operation, and an identity with respect to this operation. Examples are: groups, rings with respect to multiplication, endomorphisms of an object..

The reasons why I think group theory is so successful are several:

  1. Its theory has deep roots in Galois theory, which has shown some very advanced mathematics for its time;
  2. It is at the midpoint of a wide variety of approaches and applications: algebraic groups, invariant theory, classifying groups in algebraic topology, K-theory in homological algebra, representation theory in linear algebra and lot more, differential geometry with Lie groups, connections with (semisimple) Lie algebras, combinatorics in finite group theory, graph theory in the study of free geoups... It's really incredible.
  3. All such quoted theories are surprisingly well understood and shows rich but not impossible structures.

However - there is always an however in good stories - sometimes it's us human that needs additional hypothesis to better study and understand things. That's why, also in the case of finite groups, many people for example will concentrate on p groups, and so on... Maybe, that's why we concentrate on groups and not on general monoids, isn't it? But let's take a step back. What we are doing? What we want from an algebraic structure? Where do they generate?

Let me introduce you the concept of Operad. An Operad is a collection of sets $O(n) $, that you should think as corollas with $n$ leaves directly connected to a root below. Such operad comes equipped with a way of composing such corollas

$$\theta: O(n_1) \times \ldots \times O(n_k) \times O(k) \to O(n_1 + \ldots + n_k) $$

Here you have to think that you are "plugging in" the k corollas with respectively $n_1, \ldots, n_k$ leaves into a corolla of k leaves, to get a corolla of $n_1+\ldots+n_k$ leaves. This "composition of corollas", when you plug in three layers, should be somehow associative. Doesn't matter however.

Let me give you an example of an operad. This is very important. Take an object X in a monoids category. If you are not comfortable with this, you can imagine this is a set, a vector space, a topological space... The corresponding notion of endomorphism will vary as you imagine. Monoidal here roughly means that you can do the direct product, and let's assume that it is a coincidence with our monoids for the moment.

Define the endomorphism operad $End_X$ as having $End_X(n) = Hom(X^n, X) $, and the composition of corollas is given by

$$ \theta(f_1, \ldots, f_k, g) (x_1, \ldots, x_k) = g( f_1(x_1), \ldots, f_k(x_k)) $$

Where $f_i: X^{n_i}\to X$, $x_i \in X ^{n_i}, g: X^k \to X$. This is a bit difficult to digest, but you can imagine you have plugged in your inputs $x_i$ in the corollas (which are the $f_i$ and $g$) and let them flow to the root.

Let me give you another two examples. Define $Ass$ to be the operad that has in degree $n$ the symmetric group $S_n$ and the composition is given in a weird way, but you will understand it in a moment. Note that $S_n$ is in bijective correspondence with the possible ways of ordering $n$ variables in a product. Define $Comm$ to have just one element in any degree.

A last boring definition. A morphism of operads $O \to P$ is a degreewise map $O(n) \to P(n) $ that commutes with corollas composition. Suppose now you have a morphism of operad $Comm\to End_X$ for some $X$. What does this stands for? Well, for example, the image of the unique element in $Comm(0)$ defines an element in $e \in Hom(X^0,X)\simeq X$. It also chooses an element $\mu \in Hom(X^2, X)$, by taking the image of the only element in $Comm(2) $. If you check carefully the commutativity relations of an operad morphism, you will discover that this $\mu$ should define a commutative and associative product on $X$, and that $e$ should be the neutral element wrt $\mu$. Conversely, a commutative and associative structure on $X$ with a neutral element defines such a morphism. Cool, isn't it? You can do the same trick with $Ass$, and you will discover that he is responsible for the creation of associative products.

In analogous ways, you can produce a wide variety of operads that encode several algebraic structures. Monoids, for examples, are encoded by $Ass$. There also Lie Algebras, Rings, Hopf Algebras, Little disk Algebras, Gerstenhaber Algebras, Poisson Algebras... There is also a great feature of this way of crafting algebraic structures: relatioms that holds with an equal can be relaxed to hold with an "homotopic to", and all homotopies appearing in this way can be organized into an higher structure. Infinity category deals with formalizing this stuff. That's a bit of an " adult" observation, but I swear you it can be of concrete importance (it is for example in my present work).

But... Operads can't encode groups! You can just say that are special kind of monoids, but there isn't any operad "Grps" encoding the axioms of a group. This is not my main argument, though; the main thing is that all the construction we made relies upon monoidal categories, which are fundamental to all of the factory of algebrsic structures. And yes, a symmetric monoidal category is in first approximation exactly a monoid in which we use categories to encode the axioms: is a category $C$ with a functor $ \mu: C \times C \to C$ that is associative, and with a "neutral element", i.e. an element $1 \in C$ such that $\mu(X, 1) \simeq X \simeq \mu(1, X) $. There is some relatioms that this isomorphisms should satisfy (penthagon relations I think), but this is a technicality). In Lurie treatment of Infinity categories, which is at the boundary of algebra and topology, one cannot live without symmetric monoidal categories (but could live without group theory).

Sorry for the lengthy answer. Please comment and let me know what you think! I am myself doubtful about this monoid-groups battle... I think I'll opt for polygamy in the end :)


An arbitrary semigroup may lack an identity (or left or right element) element and/or inverse elemments for every element in the semigroup. This prevents such a rich theory without other constraints on the semigroup.

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    $\begingroup$ This doesn't really suffice as an answer. For example I could just as easily argue that because a group lacks the second operation and the extra properties a ring has, that group theory is prevented from being as rich as ring theory (which is clearly not the case). $\endgroup$ Jan 15, 2013 at 17:58

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