# Does invertability and closure imply identity?

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?

• What alternate axiom(s) are you proposing, precisely?
– lulu
Dec 30, 2018 at 18:01
• @lulu none, I was wondering why we can't remove this axiom because it appears to already be implied. I think I get it now though. Dec 30, 2018 at 18:18
• Because you need that $\, ae=ea=a\,$ for all $\,a.$ Dec 30, 2018 at 19:04
• It's worth noting that from the perspective of varieties (of algebraic structures), the only way to enforce that an algebraic structure not be empty is to demand that there be something in it - the identity is often the only constant specified in a algebraic structure and the only thing preventing the empty set from satisfying some axioms. Dec 30, 2018 at 21:50

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

• +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, Dec 30, 2018 at 18:04
• (And if both $xa=b$ and $ax=b$ always have solutions, then you automatically have identities and inverses). Dec 30, 2018 at 18:08
• 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. Dec 30, 2018 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. Dec 30, 2018 at 18:43
• 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? Dec 30, 2018 at 20:45

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.

• $e_a e_b=e_b=e_a$
– user608030
Dec 30, 2018 at 18:18

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 :)