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Sorry if this is a basic question or I'm overthinking it, but if an algebraic structure has inverse elements (or at least for a member $a$), that means $a^{-1}a=e$, and if there's closure then e is an element of the set. So in the case of defining a group, for instance, why do we need to include identity? Is it already implied?

José Carlos Santos
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Pineapple Fish
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3 Answers3

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Because if we were not assuming the existence of the identity element $e$, we would not be able to express the idea that $a^{-1}a=e$.

José Carlos Santos
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    +1 One can express the idea that $xa=b$ has a solution for all $a$ and $b$, though. Together with identities (and associativity) this is equivalent to having left inverses, but it makes sense to speak of even in structures without identities, – hmakholm left over Monica Dec 30 '18 at 18:04
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    (And if both $xa=b$ and $ax=b$ always have solutions, then you automatically have identities and inverses). – hmakholm left over Monica Dec 30 '18 at 18:08
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    To check my understanding, while invertability does require identity it adds something extra, so it's worth specifying. But indentity is a basis for invertability though so it too has to be specified... I suppose I'm still a bit confused though since while you do need identity for invertability I don't see why you have to say it explicitly as an axiom because it's not adding anything new. – Pineapple Fish Dec 30 '18 at 18:37
  • If you assert that it adds nothing new, please explan how to express the existence of an inverse of each element without it. Unless that the existence of an inverse becomes$$(\forall a\in G)(\exists a'\in G)(\forall b\in G):a'ab=ba'a=b,$$which is a bit heavy in my opinion. – José Carlos Santos Dec 30 '18 at 18:43
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    What I'm trying to get at isn't identity existing for invertability, but rather that it seems that inverse is a sort of "compound" statement, asserting both that inverse elements exist but also with this statement you can conclude there has to be an identity, because without it, as you said, you couldn't even define invertability. In your answer you said WHY identity has to be there but what I'm asking about is if it has to be there in one way WHY add it in a second way? It feels unnessesary to be redundant, introducing this as its own new and seperate axiom.I hope I'm explaining myself well? – Pineapple Fish Dec 30 '18 at 20:45
  • No, not really. Which axiom are you claiming to be redundant? – José Carlos Santos Dec 30 '18 at 20:48
  • @BenjaminThoburn The intuition of an inverse element is: "If I do $a$, there's some other group operation that lets me undo $a$." That means that if you do $a$, then undo $a$, you have done nothing at all. But what does it mean to say "$g$ does nothing"? That means that $gh = h$ for all other $h$. So we assert that such an element exists. But even though you need one axiom to state another, that doesn't mean it's a good idea to roll them into one axiom: they are stating two very different ideas. –  Dec 30 '18 at 21:03
  • @ José Carlos Santos sorry. :( Is there a specific thing that I wrote poorly? Identity. And not that it doesn't have to be there, because it does, but that we 1) say it as an axiom but also 2) – Pineapple Fish Dec 30 '18 at 22:06
  • *say it as an axiom by itself like one of the four group ones while we 2) imply it again by saying we have inverse elements. Here's how; conditional: if you don't define an identity you can't define inverses (as you said). Contrapositive: if you can define inverses then you do define an identity. By defining inverses you've made a 2 in 1 statement, that is to say you had to have defined an identity. But then you tell them, AGAIN (for a second time), to define an identity. I agree the statement in your answer is true, but I don't think it really answers the question. I am not saying you cont'd – Pineapple Fish Dec 30 '18 at 22:28
  • don't have to define the identity, but am wondering why we needlessly make identity it's OWN axiom (sorry I keep using capitals but I can't do italics). Why not shorten the group definition to just say groups require associativity, inverse, and closure because inverse requires identity also. – Pineapple Fish Dec 30 '18 at 22:34
  • Please state which axioms you propose, then, I don't see how you want to have an inverse of each element without an identity element. – José Carlos Santos Dec 30 '18 at 22:39
  • I do think there has two be the identity axiom, I just don't understand why it has to be stated if you're already saying there's inverses (and closure). Here's an example: imagine you're tasked with checking that a set has the three properties closure, inverse, and identity, in this order. (1) It has closure. (2) Now you make sure each element has an inverse, but have to first find the identity. (3) find the identity (which you've already had to do with (2)). So why not cut out (3). (Cont'd) also should this be moved to chat or sumthing? – Pineapple Fish Dec 31 '18 at 09:07
  • Alternativly, if I give you a set, and tell you, as axiomatic, that every element has an inverse, you know you'll be able to show that an identity element exists. So yes "Because if we were not assuming the existence of the identity element e, we would not be able to express the idea that $a^{−1}a=e$." But I think the source of (my) confusion may be you saying that to create inverse you must first create identity. I'm using given instead of create; if you're given existence of an inverse your given existence of an identity. And, since axioms are supposed to be givens (if I'm correct)... – Pineapple Fish Dec 31 '18 at 09:21
  • I have nothing to add to what I wrote before. – José Carlos Santos Dec 31 '18 at 09:27
  • And you assert in a group for example if invertability (and closure) hold, and this implies identity so I only need to say explicitly (that's the key word) that groups need closure, inverse, and associativity, right? – Pineapple Fish Dec 31 '18 at 09:52
  • No, I did not assert that and, as I wrote, I have nothing to add to what I wrote before. – José Carlos Santos Dec 31 '18 at 09:57
  • Let us continue this discussion in chat. – Pineapple Fish Dec 31 '18 at 15:49
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It's a reasonable question whether identity needs to be its own axiom. The inverse axiom requires that $a^{-1}$ exist for each $a$ such that $a^{-1} a=a a^{-1} = e$, which of course uses the identity in the definition. Instead, you might extend the inverse axiom as follows:

For each $a\in G$, there exists an element $a^{-1}\in G$, such that $a^{-1}a=aa^{-1}$ (call the result $e_a$), and such that for all $b\in G$, $e_a b=be_a =b$.

This seems to be weaker, allowing multiple distinct identity-ish elements, but it's not hard to show that they must actually all be the same element. So extending the inverse axiom in this way gets you back to the same definition of a group, except for one thing: without the identity axiom, the empty set would vacuously constitute a group.

mjqxxxx
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If we leave out the identity axiom, the group would not be necessarily nonempty. Then it would allow us to write e.g. the group $\langle a a^{-1} \rangle$ as $\emptyset$ because nothing to be done implies we don't need any set at all. If we define $\emptyset$ as an element, then that is just exactly the identity axiom represented by a different representation.

Without at least one identity as an element of a group, e.g. even the coset theorem would fail. Coset theorem needs us to first have $1 \in N$ where $N \leq G$, thus $g = g1 \in gN$, in order to show that:

$G = \bigcup\limits_{g \in G} gN$,

which is otherwise the set $N$ itself would not be part of the group. Heck, the operation $\emptyset N$ is not even defined! If it is defined as $N$, then it would be the identity axiom all over again.

We can of course, define the identity axiom as part of the inverse axiom, but then it would not be different than by first placing the identity axiom just before the inverse axiom. Guess the later would look more elegant :)

Ari Royce Hidayat
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