Need help confirming which algebraic structure this is

Question

Let's define a binary operation $*$ on $\mathbb{R}$ such as $$ a * b = e^{a+b} $$ and investigate which algebraic structure this is.

Well first of all we notice that the operation is closed under $\mathbb{R}$ since if $a,b \in \mathbb{R}$ and $e^x$ if defined for entire $\mathbb{R}$ then $e^{a+b} \in \mathbb{R}$. From the fact that $\mathbb{R}$ is a field we get commutativity. This is where I start to doubt myself a little. It feels like we do not have associativity since $$ (a * b) * c = e^{a+b} * c = e^{e^{a+b}+c} \neq e^{a+e^{b+c}} = a * e^{b+c} = a * (b * c)$$ I also don't think there exists an universal identity element $x$. Only way I can think of getting $a * x= a$ is when $x = ln(a)-a$, then $a * x = e^{a+ln(a)-a} = e^{ln(a)} = a$, but this is not universal or defined for all $\mathbb{R}$. Since we do not have an identity element, we cannot talk about an inverse either. Is this correct or am I going wrong here? If this is true, what algebraic structure this is? Commutative groupoid?

"the operation is closed under $\bf R$...." No; $\bf R$ is closed under the operation. — Gerry Myerson, Mar 14 '19 at 06:22

score 2 · Accepted Answer · answered Mar 14 '19 at 06:24

It is easy to check the operation is not associative: $$\begin{aligned} (0 * 0) * 1 &= 1 * 1 = e^2 \\ 0 * (0 * 1) &= 0 * e = e^e \end{aligned}$$ You've already observed there is no unique identity element.

It is commutative, and also satisfies some bizzare properties like $a * (b + c) = b * (a + c) = c * (a + b)$. But perhaps without further observations, it is just a commutative magma.

José Carlos Santos · Answer 2 · 2019-03-14T06:42:25.983

1

You are right; there is no identity element. If $a$ was such element, we would have $a*a=a$, but $a*a=e^{2a}>a$. So, yes, what we have here is a commutative groupoid.

edited Mar 14 '19 at 06:42

answered Mar 14 '19 at 06:25

José Carlos Santos

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Surely anything with “group” in the name must be associative. – Joppy Mar 14 '19 at 06:34
Not at all. – José Carlos Santos Mar 14 '19 at 06:42
Interesting, there is a different definition here: https://en.wikipedia.org/wiki/Groupoid – Joppy Mar 14 '19 at 06:44
That's about the definition of groupoid in the context of Category Theory. – José Carlos Santos Mar 14 '19 at 06:46
I see that, but it also defines a groupoid as an algebraic structure with a (partial) operation obeying various axioms. I’ve never heard a groupoid mean the same thing as a magma before, learn something new every day! – Joppy Mar 14 '19 at 06:48

Need help confirming which algebraic structure this is

2 Answers2