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I have come across an exercise that I cannot solve and honestly, to me, it doesn’t even seem true.

It goes like this:

Let $G$ be a non empty set and $*$ an associative operation on $G$ such that:

  • $\exists e\in G:\forall x\in G: x*e = x$ (right identity)
  • $\forall x\in G: \exists y\in G: x*y = e$ (right inverse)

Prove that $(G,*)$ is a group. (Hint: start by proving that $g*g = g \implies g = e$).

I tried to solve this in the following way:

$g*g = g \implies (g*g)*r = g*r \implies g*(g*r) = e \implies g*e = e \implies g = e$ where $r$ is a right inverse of g. We can actually see that the statements $g*g = g$ and $g = e$ are equivalent.

After this I tried to prove that the statements implied the existence of a left identity and its equality with $e$, but I had no success so far.

Hope you can help and thank you in advance.

Duarte Costa
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  • I suggest you to change the title to a clearer statement like "Is the existence of right inverses and a right identity sufficient for $G$ to be a group?" or something like that. edit: Thanks. I had already upvoted your question. – stressed out Jan 24 '19 at 12:56
  • As a side note, $e_l$ (a left identity, if one exists) itself has a right inverse. This implies that $e_l = e$. – stressed out Jan 24 '19 at 13:01
  • Thank you, I appreciate the help from both – Duarte Costa Jan 24 '19 at 13:15

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