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In "A First Course in Abstract Algebra", Edition 7, p.43, Fraleigh writes that

It is possible to give axioms for a group $\left<G,*\right>$ that seem at first glance to be weaker, namely:

  1. The binary operation $*$ on $G$ is associative.
  2. There exists a left identity element $e$ in $G$ such that $e*x = x$ for all $x \in G$.
  3. For each $a \in G$ there exists a left inverse $a'$ in $G$ such that $a' * a = e$

From this one-sided definition, one can prove that the left identity element is also a right identity element, and a left inverse is also a right inverse for the same element. Thus these axioms should not be weaker, since they result in exactly the same structures being called groups. It is conceivable that it might be easier in some cases to check these left axioms that to check our two-sided axioms. Of course, by symmetry it is clear that there are also right axioms for a group.

Does the above few axioms assume that the right identity element exists in the first place?

Consider $a * b = \left|a\right|b$. There are at least two possible solutions for a "right identity element", namely:

$$-1 * x = -1 \implies x = -1$$ $$1 * x = 1 \implies x = 1$$

Even though the group with operation $*$ satisfies axiom 2 of the weaker definition, it seems that $\left<G,*\right>$ cannot be a group, because using the original axioms of a group, there is no identity element $e$ such that $e*a = a * e = a$ for all $a \in G$.

How then do the above axioms "result in exactly the same structures being called groups"?

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  • $\begingroup$ Your example does not satisfy the first axiom. It's necessary to prove that $x * e = x$ for $e$ - left identity element. You do NOT have to assume there is such a thing as a RIGHT identity element. $\endgroup$ – Lurco Jul 31 '15 at 14:05
  • $\begingroup$ Hi @Lurco, thanks for the reply. On one hand $a*(b*c) = a*(|b|c) = |a||b|c$ while on the other, $(a*b)*c = (|a|b)*c = ||a|b|c = |a||b|c$. Both expressions are equal, so it can indeed be verified that the first axiom (associativity) is satisfied. $\endgroup$ – Yiyuan Lee Jul 31 '15 at 14:15
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This is a classical exercise in group theory.

Clearly you have to prove that the left identity $e$ is also a right identity and that every left-inverse is also a right inverse.

Now consider the equation $$(x''*x')*x=x''*(x'*x)$$ by the axioms we have that $$(x''*x')*x=e*x=x$$ on the other hand $$x''*(x'*x)=x''*e\ .$$ So $x=x''*e$ and multiplying by $x'$ on the right we get $$x*x'=(x''*e)*x'=x''*(e*x')=x''*x'=e$$ hence every left inverse is also a right inverse.

For $x$ generic element we have also that $$x*e=x*(x'*x)=(x*x')*x=e*x=x$$ which proves that $e$ is also a right identity.

Since both the right-identity and right-inverse axioms from groups follows from these weaker axioms it follows that every model of these weaker axioms is a group.

Edit: A different way to prove that $e$ is also a right identity is the following: starting from $x=x''*e$ you can multiply on the right by $e$ obtaining $$x*e=(x''*e)*e=x''*(e*e)=x''*e=x\ .$$

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Both possibilities (that the identity is $-1$ or $1$) are not coherent with your definition of multiplication and the requirements of the theorem.

The identity is $1$. Then the element $-1$ does not have a left inverse in that situation.

For there to be a left inverse (when the identity is $1$), there needs to be $a' \in G$ s.t. $$a'*(-1) = 1$$ This means that $$|a'|(-1) = 1$$ which has no solutions.

The identity is $-1$. Then the element $1$ does not have a left inverse in that situation.

For there to be a left inverse (when the identity is $-1$), there needs to be $a' \in G$ s.t. $$a'*1 = -1$$ This means that $$|a'|1 = -1$$ which has no solutions.

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