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Given that (i) $ae = a$ and (ii) $aa^{-1} = e$, how can you show, without establishing commutativity as an axiom, that $ea = a$ and that $a^{-1}a = e$? (I guess we would in a sense be proving commutativity, which I'm having difficulty with considering we consider it as a group axiom.)

To show $ea = a$, I tried showing $ea = aa^{-1} ae = ea = e$ (from i and ii). I am pretty certain this approach is wrong.

EDIT: $e$ is the identity element and $a$ is an element in the set.

EDIT2: I edited my question to be more specific; I'm trying to show commutativity of the identity element with an arbitrary element in the same set and also the commutativity of a and its inverse if it exists in the same set.

zipirovich
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Dominated Convergence Theorem
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  • Does the symbol $e$ have any special significance in the group... or is it just an arbitrary element? Are you quantifying over all $a$, or are you just talking about one element $a$? – rschwieb Oct 11 '17 at 18:15
  • Just a comment: proving that $ea=ae$ does not imply commutativity, which states that $ab=ba$ for all $a,b$ in the group. $ea=ae$ holds for all groups where $e$ is the identity, but the statement that it holds for all $b$ is stronger. – John Gowers Oct 11 '17 at 18:16
  • The definition of $a^{-1}$ is that $aa^{-1}=a^{-1}a=e$, so it would be bad to use that symbol as a "maybe only one-sided inverse". Similarly, the definition of the identity is that $ae=ea=a$ for all $a$. I don't see why that's an issue... – rschwieb Oct 11 '17 at 18:20
  • Both of these properties are contained in the axioms of a group, so there is nothing to prove unless you are assuming something less than the axioms of a group to begin with. – rschwieb Oct 11 '17 at 18:25
  • I'm trying to deduce that ea=ae=a from the stated conditions, not from stating it from group axioms. – Dominated Convergence Theorem Oct 11 '17 at 18:27
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    @rschwieb It's disingenuous to claim that $aa^{-1}=a^{-1}a=e$ and $ae=ea=a$ are contained in the axioms of a group and that there is therefore nothing to prove. For simplicity, they are usually included, but there are alternative equivalent axiomatizations of group theory, including the one where the identity and inverse axioms are spelled as $ae=a$ and $aa^{-1}=e$: since the other two equations can be proved from these two in the presence of associativity, they do not need to be given. – John Gowers Oct 11 '17 at 18:28
  • I guess it's more accurate to say that you have a right identity (not just an identity) and right inverses for all elements, and then you want to prove that it's actually a two-sided identity and that inverses are two-sided two. Assuming associativity, this is true. As a fun fact, note that right identity together with left inverses wouldn't work, as far as I remember. – zipirovich Oct 11 '17 at 19:08
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    The proof I first saw had outline: first prove $a \cdot a = a \Rightarrow a = e$ (by multiplying both sides on the right by $a^{-1}$); then apply this to the case $a^{-1} a$ to show $a^{-1} a = e$; then finish off by $ea = (a a^{-1}) a = a (a^{-1} a) = ae = a$. – Daniel Schepler Oct 11 '17 at 19:12
  • Dear @JohnGowers : I know what you mean, but saying I am being disingenuous is uncalled for. What group axioms am I supposed to assume aside from the usual ones? Things will get clearer once the assumptions being made are cleared up. – rschwieb Oct 11 '17 at 19:14
  • @Primes What axioms would you like to assume to define a group? When I asked earlier "are you quantifying over all $a$" I meant "are you saying For all $a$, $ae=a$ and $aa^{-1}=e$" It's unclear whether or not you are talking about a single $a$ or you are assuming it for everything in the set. – rschwieb Oct 11 '17 at 19:19
  • @zipirovich If that's the case, then this is going to be a duplicate. – rschwieb Oct 11 '17 at 19:21

1 Answers1

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EDIT: Here I'm assuming associativity. I don't think this is true without it.

Proposition 1. $aa^{-1}=e$ implies $a^{-1}a=e$.

Proof. $$ e=(a^{-1})(a^{-1})^{-1}=(a^{-1}\cdot e)(a^{-1})^{-1}=a^{-1}\cdot (e\cdot (a^{-1})^{-1})$$ $$ =a^{-1}((a\cdot a^{-1})\cdot (a^{-1})^{-1}) =a^{-1}\cdot a\cdot e=a^{-1}\cdot a.$$ Proposition 2. If $ae=a$ for all $a\in G$ then $ea=a$.

Proof. $$ a\cdot e=a\cdot (a^{-1}\cdot a)=(a\cdot a^{-1})\cdot a=e\cdot a=a.$$

Corollary. $ae=ea$ for all $a\in G$.

Alekos Robotis
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