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I'm starting a course on algebraic structures in university, and this is one of the first problems. I want to know if my solution is valid. The question goes like this:

Being $G$ non-empty semigroup (set with associative binary product). We know that $G$ verifies that:

  • $\exists e\in G:ex=x \forall x\in G$ ("left" neutral element).
  • $\forall x \in G, \exists y\in G : yx=e$ ("left" inverse element).

Prove that $G$ is group.

My solution goes like this: $$e=ee=(g^{-1}g)e=g^{-1}(ge)$$ Then necessarily $ge=g$, because $g^{-1}(ge)$ must be equal to $e$. So that proves that $e$ is in fact left and right neutral element of $G$.

Now, we see that: $$g=ge=g(g^{-1}g)=(gg^{-1})g$$ and we conclude that $gg^{-1}=e$ because $(gg^{-1})g$ must be equal to $g$, so we conclude that $g^{-1}$ is left and right inverse element of $g$.

Given this, I conclude that $G$ is in fact group.

Is my reasoning correct? I'll thank any help.

Shaun
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Alejandro Bergasa Alonso
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    Using the notation $g^{-1}$ before proving that there is a unique two-sided inverse is not a good idea. – Christoph Oct 25 '20 at 13:12
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    It looks like you are using uniqueness of right inverses when you have $e=g^{-1}(ge)$ and conclude $ge=g$, aren't you? – Christoph Oct 25 '20 at 13:14
  • @Christoph assume i wrote $y$ instead of $g^{-1}$. The proof is the same and is valid right? It's just notation – Alejandro Bergasa Alonso Oct 25 '20 at 16:33
  • As I said: how did you get $ge=g$? I'm not even sure the task is properly stated: what is $e$ in the second condition, when the first condition doesn't assert uniqueness in "$\exists e$"? – Christoph Oct 25 '20 at 17:02

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