Does invertability and closure imply identity?

Question

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?

Answer 1

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

Answer 2

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.

Answer 3

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 :)