Why is associativity required for groups?

I'm doing a linear algebra paper and we're focusing on groups at the moment, specifically proving whether something is or is not a group. There are four axioms:

  1. The set is closed under the operation.
  2. The operation is associative.
  3. The exists and identity in the group.
  4. Each element in the group has an inverse which is also in the group.

Why does the operation need to be associative?



5 Answers 5


It is not that associativity is required for groups... That is quite backwards: the truth is actually that groups are associative.

Your question seems to come from the idea that people decided how to define groups and then began to study them and find them interesting. In reality, it happened the other way around: people had studied groups way before actually someone gave a definition. When a definition was agreed upon, people looked at the groups they had at hand and saw that they happened to be associative (and that that was a useful piece of information about them when working with them) so that got included in the definition.

If I may say so, it is this which is important to understand. The way we teach abstract algebra nowdays somewhat obscures this fact, but this is how essentially everything comes to be.

  • 1
    $\begingroup$ A very true comment! $\endgroup$
    – David Ward
    Jul 19, 2012 at 8:05
  • $\begingroup$ If I remember correctly, there were a few definitions of "group" just over a hundred years ago. It was in Burnside's time that it was fixed down (I mean, fundamental groups weren't a "thing" until this time, so this all makes sense.) $\endgroup$
    – user1729
    Jul 22, 2012 at 15:49
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    $\begingroup$ @user1729 The original definition of a group-I believe by Cayley (?) -was of a group of transformations, what today are called in geometry and physics, the dihedral groups in the plane and three dimensional space. The first abstract definition of a group using the familiar axioms was due to Leopold Kronecker, I believe in 1879. However, Kronecker's definition wasn't quite the modern one because it required commutativity-what Kronecker called a group was actually what we today call an Abelian group. It was indeed Burnside and his contemporaries that formulated the current definition. $\endgroup$ Jul 29, 2012 at 4:44
  • $\begingroup$ Very good point there $\mathbb{J}$ $\endgroup$ Oct 27, 2016 at 1:07

Groups are an abstraction. What do they abstract? The idea of symmetry. Symmetries are functions from a set to itself that preserve some structure of that set; for example, the symmetries of a square are rotations and reflections, and they preserve "squareness" (to put it vaguely).

The multiplication in a group abstracts composition of symmetries (for example "rotate $90^{\circ}$, then reflect about the line $x = y$"), and composition of functions is always associative.


The formalist's answer is: it is just a definition. You could just as well consider studying algebraic structures that satisfy all the axioms for a group except for associativity, and you would be then studying loops.

Now the question might be: why is the study of groups more ubiquitous than the study of loops? There are historical reasons (surely others with greater knowledge can expand upon this), and the fact that most loops that arise naturally when doing math are in fact groups is probably a reason too.

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    $\begingroup$ Thanks for the answer, I have one more question if you don't mind. Because of the associativity axiom we obviously need to check if the operation is associative. Does the associativity of the operation always "come from" the underlying set? How do you go about proving something is associative without prior knowledge that it is? For instance, how would one prove that $1 + (2 + 3)$ = $(1 + 2) + 3$ without simply accepting that it is? This has been confusing me because all proofs we do for associativity generally just state that it is. $\endgroup$ Jul 19, 2012 at 4:14
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    $\begingroup$ @user1520427: for example, I know that $2 + 3 = 5, 1 + 5 = 6$. I also know that $1 + 2 = 3, 3 + 3 = 6$. So I can verify associativity in this case. If you want to prove that integer addition is associative you would need to start from a particular definition of integer addition. $\endgroup$ Jul 19, 2012 at 4:44
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    $\begingroup$ This is a good question! Generally, proving associativity of an operation is either very easy, because it comes immediately from the underlying structure, or it is very difficult/tedious, because you have to verify it for a large number of cases. This second situation can arise if you are trying to find all groups of a given order - particularly when the order is a power of a prime. $\endgroup$
    – Derek Holt
    Jul 19, 2012 at 9:21
  • $\begingroup$ What historical reasons, can you mention one? $\endgroup$ Mar 16, 2017 at 18:37

In short, because that's how we choose to define them, because adding associativity allows us to study certain things more robustly.

There are algebraic structures that are group-like but don't satisfy all those axioms. A quasi-group need not be associative, and a loop need not be associative, but must have unity.

So, it may sound circular, but a group must be associative because if it is not, then it is not a group. A more apt question may be "why do we study group theory and not quasi-group theory?" In some senses, associativity gives us more freedom and power.

  • 1
    $\begingroup$ So nice, never thought like that. Wished you could express in more details or give reference to how associativity gives us more freedom & power. If you wish, can frame a new question for that. But, lack proper knowledge to attract responses to that. So, will wait for your affirmation. Else, you could give here more inputs for enabling me for framing such a question. As for me, I take other view to the question by considering that groups have to be associative as then reversibility of operations is possible. Thanks in anticipation for your kind response. $\endgroup$
    – jiten
    Oct 31, 2020 at 9:38

The important thing to understand is that fulfilling these 4 axioms is all there is to being a group-in mathematics, we construct objects by imposing conditions on them and the underlying aggregates. (I'm not using the term "set" since mathematics doesn't always deal with sets,but you get the idea.) What allows us to build a theory around a particuar defined object is the fact that whatever we choose the properties to be,those properties must be consistent with each other. It's the consequences of those properties cooexisting consistently that gives any object it's distinctive properties spelled out in the body of theorums and corollaries. And that's the answer to your question: It turns out if you don't have associativity-that is, if axiom (2) is false, then axiom (4) is false because then you can have a set which has left inverses which are not right inverses and the definition assumes the inverse ( as well as the identity) is 2 sided. Consider an element x of a group G where there is a unique identity e and a left inverse l and a right inverse r for x. Then by axiom (4),they must be equal since both yield the unique identity and every inverse must be 2 sided. But:

l = l*e = 1*( x*r) = (1*x)*r = e*r= r.

But clearly the fact of l= r is dependent on the associativity of the operation. So whatever this algebraic structure is,it's not a group without it.

Hope that answered your question.

  • $\begingroup$ I see my fan club is back and busy randomly and anonymously downvoting me again. Have at it. $\endgroup$ Jul 24, 2012 at 8:29
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    $\begingroup$ Upvoted to even things out. I just wanted to note that it's strange to call axioms "false". Moreover, the last part of this is a bit strange, since you are saying that (2) "false" implies (4) "false", but in the case that we exclude (2), we will just get structures which satisfy (1), (3), and (4) --- the ordinary groups won't necessarily be part of these structures. This doesn't mean, though, that axiom (4) is false in some way, just that we must restrict our structures. $\endgroup$
    – user2959
    Jul 28, 2012 at 22:32
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    $\begingroup$ @james It really depends on how both axiom (3) and (4) are interpreted.I assumed,as most algebraicists do, that an inverse for specific element and the identity for the entire group are 2 sided. We can weaken (3) and (4) so that the inverses and identities are assumed strictly one sided-indeed,we can assume there is AT LEAST ONE of each,which means it is not required the identity or inverses are 2 sided,let alone unique.But if we assumed this without associativity.But again,the resulting structures would not be groups unless we did a radical redefining and rederiving of their properties. $\endgroup$ Jul 29, 2012 at 4:59
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    $\begingroup$ I'm not really talking about the specific axioms here, I'm just talking in general --- axioms can't really be "false". If we don't include a particular axiom somewhere, we'll get different structures. $\endgroup$
    – user2959
    Jul 29, 2012 at 19:54
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    $\begingroup$ @james We're really saying the same thing. What I tried to do with my response-that the OP really isn't getting-is to show why it MUST be so. The truth of ALL the axioms is ALL THAT IT MEANS to be a particular structure and the second we drop any of them,we have a different structure. That's what the OP didn't seem to grasp. $\endgroup$ Aug 4, 2012 at 2:36

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