Are unique right identity and left inverse proof enough for a group?

Question

An associative * on a set G with unique right identity and left inverse proof enough for it to be a group ?Also would a right identity with a unique left inverse be a group as well then with the same logic? I don't need the proof , more so an explanation of why or why not the second case would be true or not?

Answer 1

The second situation does not parse very well as written, as “inverse” refers to identity and if you have potentially multiple identities, then you don’t know what “unique inverse” refers to.

You could sharpen it to something like the following:

Let $G$ be a semigroup such that there exists $e\in G$ such that $xe=x$ for all $x$, and such that for each $f\in G$ that is a right identity and each $x\in G$ there exists a unique $y\in G$ such that $yx=f$. Is $G$ a group?

The answer is “no”. Let $G$ be any set with more than one element and define the operation on $G$ by $xy=x$ for all $x,y\in G$. Then every element of $G$ acts as a right identity, and given any $f,x\in G$, there is a unique $y\in G$ such that $yx=f$ (namely, $y=f$). However, $G$ is not a group.

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As to the original question, it has been asked before.