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Let $G$ be a set with the operation $*$ such that:

  1. $G$ is closed with respect to $*$

  2. $*$ is asociative

  3. There exists an element $e \in G$ such that $e*x=x$ for all $x\in G$

  4. Given $x \in G$, there exists $y\in G$ such that $y*x=e$

The axioms of closure and asociativity are satisfied by the hypotesis 1 and 2. Now it is left to prove that $x*e=x$ and $x*y=e$ to conclude $G$ is a group. I've made some sustitutions to try to get the commutative property on those produts but I don't get a way to achieve it. These set of suppositions are then stronger than the group definition as they can be deduced from these ones right? What shoud I compute to prove $G$ is a group?

tobiasbriones
  • 374

1 Answers1

1

To prove $x*e = x$:

From 4 we have there is a $y$ such that $y*x = e$. We also have from 4 there is a $z$ such that $z*y = e$. Thus $z*e = z*(y*x) = (z*y)*x = e*x = x$. Thus $x = z*e$. We then see that $x*e = (z*e)*e = z*(e*e) = z*e = x$.

To prove $x*y = e$:

By the above we have $x = z*e$ where $z*y = e$. Thus $x*y = (z*e)*y = z*(e*y) = z*y = e$.

Sean Haight
  • 1,368