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I'm having some trouble with the following exercise:

Let $G$ be a non-empty set and let $*:G\times G \to G$ such that:

  1. $*$ is associative
  2. $\forall a,b \in G, \exists x \in G: a * x = b$
  3. $\forall a,b \in G, \exists x \in G: x * a = b$

Prove that $(G,*)$ is a group

We already know that $*$ is associative but I'm having some trouble proving that there exists an identity element. My plan is the following:

  1. For any $a \in G$, show that if $a * x = a$ and $x'*a = a$, then $x = x'$
  2. for any $a,b \in G$, if $a * x_a = a$ and $b * x_b = b$, then $x_a = x_b$

But I'm not being able to prove 1. How can this be done?

Shaun
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656475
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  • I would try to show that $$aG=G=Ga$$ for all $a\in G$, which is an equivalent condition to $(G, *)$ being a group. – Shaun Nov 23 '21 at 11:34
  • But doesn't that just prove that the operation $*$ is closed under $G$? @Shaun? – 656475 Nov 23 '21 at 11:35
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    No. It is a theorem of semigroup theory. See here. – Shaun Nov 23 '21 at 11:36
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    Maybe a better plan: first show that $ae=a\implies be=b$. So there is a right identity element $e$. By the same token there is a left identity element, which must be equal to $e$. Plainly, each element $x$ has a right inverse $x'$. Finally show that $x'$ is also a left inverse: $x'x=x'xe=x'xx'x''=x'x''=e$. – bof Nov 23 '21 at 11:54
  • I'll give that a try @bof – 656475 Nov 23 '21 at 12:06

1 Answers1

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Let $g$ be any element of $G$ and take $a = b = g$ in assumption 2.
It follows that there exists $e \in G$ such that $g * e = g$.

It follows that $h*g*e = h*g$ for any $h\in G$.
By assumption 3, any element of $G$ can be written as $h*g$ for some $h \in G$.
Thus we have $x*e = x$ for any $x \in G$.

This shows that there exists a right identity $e$ of $G$.
The same proof, with left and right switched, shows that there exists a left identity $e'$ of $G$.

We then have $e' = e'e = e$, which means that $e$ is an identity of $G$.

It remains to show that every element has an inverse.
For any $g \in G$, assumptions 2 and 3 say that there exist $h$ and $h'$ such that $g*h = e = h'*g$.
As $h' = h'*e = h'*g*h = e*h = h$, we see that $h$ is an inverse of $g$.

WhatsUp
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