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Consider the set of real numbers $\mathbb{R}$, which we know is the only complete ordered field up to isomorphism.

If we fix the following axioms for the 'addition' operation $\bot:\mathbb{R}^2\to \mathbb{R}$:

  1. Associativity
  2. Commutativity
  3. Identity element 0 (not that there exists some 0 element, but that the identity element is actually the canonical 0).
  4. All elements have their own unique inverse

[this defines an abelian group over the set $\mathbb{R}$ with identity 0]

and finally

  1. $\bot$ is distributive over multiplication: $\forall a,b,\lambda\in \mathbb{R}, $ $\lambda\cdot (a\bot b)=(\lambda \cdot a) \bot (\lambda \cdot b).$

Do these axioms uniquely determine addition on $\mathbb{R}$ up to automorphism?

The example of $a\bot b := (a^3+b^3)^{1/3}$ shows that these axioms do not uniquely determine addition. However, the automorphism $f(x)=x^3$ makes this operation still effectively the same as addition. The question is whether all operations $\bot$ which adhere to these axioms have an automorphism with addition.

And as a secondary question, if it turns out that they do:

Is the fifth axiom necessary, or is the only abelian group with identity 0 on the real numbers addition (up to automorphism)?

EDIT: The second part has been answered, and the answer is yes, it is necessary.

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    $\begingroup$ Clearly the answer to your sceond question is yes, the fifth axiom is necessary. Otherwise, without it you're asking "are there two non isomorphic abelian groups with cardinality $\mathfrak{c}$ ?" to which the answer is obviously yes (you can simply construct a counter-example, for instance $(\Bbb{Z}/2\Bbb{Z})^{\Bbb{N}}$ which is a torsion group and therefore isn't isomorphic to $\Bbb{R}$) $\endgroup$ – Max Jun 19 '17 at 19:41
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    $\begingroup$ @Max Or even more simply, $(\mathbb{Z}/2\mathbb{Z})\times \mathbb{R}$. $\endgroup$ – Noah Schweber Jun 19 '17 at 19:42
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    $\begingroup$ @NoahSchweber why make things easy when you can make them complicated haha ... $\endgroup$ – Max Jun 19 '17 at 19:43
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    $\begingroup$ @MathTrain None of that matters: if I have a group $G$ with cardinality continuum, then there's a bijection between $G$ and $\mathbb{R}$ which sends the identity of $G$ to $0$ - this induces a group structure on $\mathbb{R}$ with the desired properties, isomorphic to $G$. You need some further conditions on the group operation, here. $\endgroup$ – Noah Schweber Jun 19 '17 at 19:51
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    $\begingroup$ Yes I know but what you're asking (without the distributivity condition) is equivalent to "Do there exist two non isomorphic abelian groups of cardinality $\mathfrak{c}$ ?" . The fact that the two questions are equivalent is relatively easy once you see that having a bijection is simply renaming things. $\endgroup$ – Max Jun 19 '17 at 19:51
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Your axioms do not determine addition uniquely. Note that if $K$ is any field and $f:\mathbb{R}^\times\to K^\times$ is an isomorphism of the multiplicative groups, then the operation $(x,y)\mapsto f^{-1}(f(x)+f(y))$ (and $(x,0)\mapsto x$, $(0,y)\mapsto y$) will satisfy your axioms, and will only be equivalent to ordinary addition in your sense if $K$ is isomorphic to $\mathbb{R}$ as a field.

Now note that the abelian group $\mathbb{R}^\times$ is not too hard to understand. It is the direct sum of the subgroups $\{\pm1\}$ and $\mathbb{R}_+$, and $\mathbb{R}_+$ is a vector space over $\mathbb{Q}$ since every positive real number has an $n$th root for any $n$. More generally, the same description holds for $K^\times$ if $K$ is any ordered field in which every positive element has an $n$th root for any $n$. Since two $\mathbb{Q}$-vector spaces of the same uncountable cardinality are isomorphic, it follows that if $K$ is any ordered field of the same cardinality as $\mathbb{R}$ in which every positive element has an $n$th root for any $n$, then $K^\times\cong\mathbb{R}^\times$.

However, such a field need not be isomorphic to $\mathbb{R}$, and so can give rise to a different "addition" on $\mathbb{R}$ satisfying your axioms. For instance, you could take $K$ to be any non-archimedean real-closed field of cardinality $2^{\aleph_0}$ (e.g., the real closure of $\mathbb{R}(x)$ where $x$ is infinitely large).

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  • $\begingroup$ I'm sorry, I don't really understand this answer. I have one year of undergrad math, and no background in group theory. Could you use less sophisticated terminology or symbols? I cannot tell whether this means yes or no $\endgroup$ – MathTrain Jun 19 '17 at 20:14
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    $\begingroup$ I've added a bit more explanation. I'm afraid that I can't imagine any answer to your question that does not involve some fairly high-powered machinery, though. $\endgroup$ – Eric Wofsey Jun 19 '17 at 20:43

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