# Inverse operation in a True Group with multiple identity elements

Let $S$ be a hypothetical algebraic set. It is a True Group, a Monoid where the elements may be inversed (not necessarily an Abelian group).

$S$ is of infinite size but only three of its members concern us: $a$ is a left identity element, $b$ is a right identity element, $c$ is a dual-identity element (both left and right identity). For argument's sake, $a$ ≠ $b$ ≠ $c$ ≠ $a$

Does this mean $S$ has different inverse functions, one for each identity element? Or can a True Group only have a single identity element?

(I'm only an armchair mathematician looking at the properties of algebraic operations so please be gentle! :)

• A group (I have never seen anyone refer to a true group before) has a unique identity element, and each element has a unique inverse. – Tobias Kildetoft Sep 21 '17 at 7:45
• It's difficult to answer this without knowing the precise definition of "true group". – Derek Holt Sep 21 '17 at 7:52
• @ Derek Holt: a Monoid where the elements may be inversed (not necessarily an Abelian group). – Smoke Sep 21 '17 at 9:24
• That is just a group (since groups are not necessarily abelian). – Tobias Kildetoft Sep 21 '17 at 9:28

$S$ is of infinite size but only three of its members concern us: $a$ is a left identity element, $b$ is a right identity element, $c$ is a dual-identity element (both left and right identity). For argument's sake, $a$ ≠ $b$ ≠ $c$ ≠ $a$.
Then you immediately have a contradiction since $c=ac=a$, and $c=cb=b$, and so $a=b=c$.
The similar thing can be said for uniqueness of inverses. If $b'$ is a left inverse of $b$, and $b''$ is a right inverse, then $b'=b'(bb'')=(b'b)b''=b''$. So when a left and right inverse exist, they are equal, and thus two-sided.