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I am not a mathematician but was dealing with the axioms linked/related to conservative group theory definitions.

In that they said that if we have a group $\left< G, \otimes\right >$ Then amongst the axioms, we encounter

  1. $\exists e \in G : \forall y \in G, e \otimes y = y \otimes e = y$
  2. $\forall y \in G, \exists y^{-1} \in G : y^{-1} \otimes y = y \otimes y^{-1} = e$
  3. The associativity axiom
  4. The closure axiom

My question is, were the axioms 1 and 2 limited to only (what we could call the limited axioms)

  1. $\exists e \in G : \forall y \in G, e \otimes y = y$
  2. $\forall y \in G, \exists y^{-1} \in G : y^{-1} \otimes y = e$

Then, we could say from property 1 and 2.

$$y^{-1} \otimes e \otimes y = y^{-1} \otimes y = e$$ $$\implies \left(y^{-1} \otimes e \right) \otimes y = e$$ $$ \texttt{Letting} \quad y^{-1} \otimes e = z$$ $$\forall y \in G, \exists z \in G : z \otimes y = e$$

From property 2, $z = y^{-1}$ because of uniqueness of the inverse element for each element in the group as a whole. Thus $y^{-1} \otimes e = y^{-1} = e \otimes y^{-1}$, the commutativity of the identity operation could be derived only using the properties of "associativity" and "uniqueness of the inverse elements" which are already present in the definition of a group.

Now once, we have derived the "commutativity of the identity operation", we could start with property 2 of the "limited axioms" $$\forall y \in G, \exists y^{-1} \in G : y^{-1} \otimes y = e$$ $$ \implies y \otimes (y^{-1} \otimes y) = y \otimes e = y$$ $$\implies (y \otimes y^{-1}) \otimes y = y$$ $$\implies \texttt{iff} z = y\otimes y^{-1} \implies z \otimes y = y$$ $$\implies \exists z \in G, \forall y \in G, : z\otimes y = y $$ From property 1 of the "limited axioms", $z = e$, because of the uniqueness of $e$.

Thus $\implies y \otimes y^{-1} = e = y^{-1} \otimes y$ Hence, again the property of "commutativity of inverse operators" could be derived using solely the properties of "associativity" and "uniqueness of identity elements"

Thus, my question is, do the commutativity of inverse and identity elements have to be etched into the definition of groups in group theory? Or is it a direct consequence of the "associativity of all operations" and the uniqueness of both "identity" and "inverse" elements?

Ghosal_C
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    You are using the uniqueness of the inverse in your argument, but that is proved using the usual definition of a group. If you change the axioms you can't use it unless you prove it still holds with your weaker axioms. – Captain Lama Dec 11 '21 at 13:32
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    You can see an argumentation in the question https://math.stackexchange.com/questions/433546/is-a-semigroup-g-with-left-identity-and-right-inverses-a-group (the question is different, but the OP shows what you are looking for in the question itself). – Captain Lama Dec 11 '21 at 13:51
  • conservative group theory? – anon Dec 12 '21 at 01:16
  • @runway44 "usual group theory", "run-of-the-mill group theory".......what I meant is group theory that you find in generic textbooks/pdfs – Ghosal_C Dec 12 '21 at 07:25
  • @CaptainLama I followed it. It's an interesting thread really, which takes into account same-sided inverse/ident against flip-sided inverse/ident. flip-sided inverse/ident don't really form a group. However, I will look into what you said about "uniqueness of inverse/ident. not being a fundamental property but rather a consequence of both sided inverses". Would you recommend a text on group theory to me? – Ghosal_C Dec 12 '21 at 08:06

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