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Well, I am little bit confuse in definition of group. My question is this that why we need of identity property in definition of group because if a set has a associative binary operation such that every element has inverse then it obviously contain identity element. If I'm wrong then can someone give me an example of a set which not group just because of identity property i.e. we have a set equipped with an associative binary operation along with inverse property but it does not have identity element. Please clear my doubt.

Sachin Singh
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    How do you define inverse of an element without defining identity element before? – azif00 Jun 25 '21 at 03:50
  • But if we say in inverse property that for every element a in G there exists b in G such that ab=e where e is identity of G and it has property ex=xe=x. I want to say that why we need existence of identity element while identity will belong to Group G when operation is closed and inverses exists. I have no issue with defining identity element but issue with existence. – Sachin Singh Jun 25 '21 at 04:11
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    Suppose that no identity element exists, then how do you define inverse? Given $a\in G$, what is $a^{-1}$ then? (here assume that we don't know what an identity is). – Koro Jun 25 '21 at 05:14

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Indeed it would be possible to define a group as a (nonempty) semigroup equipped with a unary function $\cdot\,^{-1}$ that satisfies $$ a(a^{-1})b = ba(a^{-1}) = (a^{-1})ab = b(a^{-1})a = b $$ for all $a$ and $b$. There are four identities because we need inverses and identities to both work from the left and right.

In fact, if we want to be even more minimalist we can get down to two identities: $$ a(a^{-1})b = b = b(a^{-1})a $$ together with associativity. This is enough to make sure that $ax=b$ and $xa=b$ always have solutions, which by this question implies that our structure is a group.

Why is it usually not done this way?

There's no real technical pressure to make the number of symbols in the signature or axiom system as small as possible. As long as everyone agrees what are and are not groups, what really matters is not how tersely you can state an axiomatic definition, but how easily you can present the definition and its usual elementary consequences in an introductory textbook -- and for that, being too parsimonious is not necessarily a win.

Additionally, doing it this way around would obscure the fact that groups are monoids that satisfy additional conditions. It can be proved, of course, but it's more instructive to have it hold immediately by definition.

Especially because beginning students of group theory are often not very mathematically sophisticated (it is often the first completely abstract definition of an algebraic structure one sees), everything that makes it harder to work with the definition is generally to be avoided.

Troposphere
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  • This is a fantastic answer. – user3716267 Jun 25 '21 at 05:17
  • It's mean we can give a alternate definition of group that a set equipped with an associative binary operation satisfying left & right cancellation law. Is it? Also, is it true only for finite set? – Sachin Singh Jun 25 '21 at 07:11
  • @SachinSingh: What I wrote in this answer is true both for finite and infinite sets. However, note that what I wrote never mentions cancellation laws. For finite sets, cancellation laws imply that $ax=b$ and $xa=b$ always have solutions, thanks to the pigeonhole principle -- but this reasoning does not work for infinite sets. Indeed, for example, addition of strictly positive integers is associative and satisfies cancellation both to the left and to the right, but has neither an identity nor inverses. – Troposphere Jun 25 '21 at 07:31
  • (I understand a "cancellation law" to be the rule that $ab=ac$ implies $b=c$, no matter whether there is a multiply-by-the-inverse argument that causes it, or the composition just happens to satisfy it anyway -- see https://en.wikipedia.org/wiki/Cancellation_property). – Troposphere Jun 25 '21 at 07:39
  • This answer was my first thought as well, but is it really an equivalent definition of groups? Specifically, what do we know about uniqueness of identity elements? – leftaroundabout Jun 25 '21 at 13:10
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    @leftaroundabout: Uniqueness of identities is a standard argument: If $e_L$ and $e_R$ are left and right identities, then $e_L=e_Le_R=e_R$, which means that they are actually a two-sided identity which means (by the same argument) must be the only identity in the structure. And the two "minimalist" axioms in this answer say exactly that $aa^{-1}$ is a left identity and $a^{-1}a$ is a right identity. [A semigroup can have several different left identities, but only if there are no right identities, or vice versa]. – Troposphere Jun 25 '21 at 13:28
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It depends what you mean by the "inverse property". You might be after the theory of inverse semigroups.

A semigroup is a set under an associative binary operation, while an inverse semigroup $S$ is a semigroup in which for every element $x \in S$ there exists a unique $y\in S$ such that $x = xyx$ and $y = yxy$ - here, $x$ and $y$ are inverse elements. See Wikipedia or these notes of Mark Lawson for more details.

Groups are inverse semigroups, but the converse is not true. For example, $\{0, 1\}$ under multiplication is an inverse semigroup which is not a group (both elements are self-inverse, so it is an inverse semigroup, but there is a $0$ so it is not a group). Another example is the following multiplication table from Wikipedia, which again defines an inverse semigroup which is not a group (multiplication is not bijective). \begin{array}{c|ccccc} &a & b & c & d & e\\\hline a& a& a& a& a& a\\ b& a& b& c& a& a\\ c& a& a& a& b& c\\ d& a& d& e& a& a\\ e& a& a& a& d& e \end{array} More examples can be found in this old Math.SE question.

An important class of examples of inverse semigroups which are not groups are the Symmetric inverse semigroups $\mathcal{I}_X$ for each set $X$; these are the "partial" bijections (that is, bijections which are not defined everywhere) on $X$ under a certain natural composition operation. These play a similar role in inverse semigroup theory as the symmetric groups play in group theory, as every inverse semigroup can be embedded into an symmetric inverse semigroup (this is the Wagner-Preston theorem).

user1729
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  • Instead of sitting down to grok the 25-element table (and convince oneself it is associative), one can also just consider the multiplication operation of any field, declaring the inverse of $0$ to be $0$, and other inverses to be the usual reciprocal. – Troposphere Jun 25 '21 at 13:37
  • @Troposphere Yes, in fact ${0, 1}$ under multiplication is the smallest example. (I've edited this one in.) – user1729 Jun 25 '21 at 13:38
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Here is what is usually done.

Whenever you encounter a binary operation, ask the following three questions:

  1. Is the binary operation commutative?

  2. Is the binary operation associative?

  3. Is there an identity element?

Now, in case the answer to the third question is "yes", you ask the fourth question.

  1. What about the inverse?

Notice that it would be completely meaningless to talk about the inverse if there is no identity.

F. A. Mala
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  • Ok I got it now that concept of inverses is directly connected with identity and that's why we have to mention it. Thanks for responding. – Sachin Singh Jun 25 '21 at 07:07