7
$\begingroup$

I am working with a group, of sorts, that contains not one, but two, binary operations. Let's call these $+$ and $*$. For either operation, the group has a common identity, $e$, and every element has an inverse.

I am able to show that these operations are associative, not only within themselves, but with each other. For example, $$(a+b)*c=a+\left(b*c\right)$$.

If this is not a group, what is the name of the type of algebra I am dealing with? Why does it seem like such groups have not received much attention?

My question is related to the one here, except that my case has every property of a group (associativity, identity, and inverse). In other words, if I disallowed either $*$ or $+$, I would still have a group.

$\endgroup$
2
  • 1
    $\begingroup$ For either operation, but $e$ must not be the same for both operations? Maybe we should call them $e_*$ and $e_+$ respectively. $\endgroup$ Commented Sep 21, 2017 at 18:05
  • $\begingroup$ Please give an example of a set with the two operations. Preferably the smallest example where the two operations are distinct. Can you do that? $\endgroup$
    – Somos
    Commented Sep 21, 2017 at 18:52

1 Answer 1

4
$\begingroup$

Let's call these $+$ and $*$. For either operation, the group has a common identity, $e$, and every element has an inverse.

Then $+=\ast$, because $a\ast c=(a+e)\ast c = a+(e\ast c)=a+c$ for every $a,c$.

$\endgroup$
7
  • $\begingroup$ Why do they need have a common identity? Can they not have different identities but still be able to fulfil the group axioms? You have proven that IF they have a common identity then the operations must be the same. $\endgroup$ Commented Sep 21, 2017 at 17:58
  • 1
    $\begingroup$ @mathreadler The part you are referring to is cut and pasted from the original question. So yes, I have "only" proven this for the case the question outlined. $\endgroup$
    – rschwieb
    Commented Sep 21, 2017 at 18:30
  • $\begingroup$ For "either" operation, not for "both". Just that there exists some element $e_1$ s.t. $e_1+g = g+e_1 = g, \forall g \in G$ and $e_2$ s.t. $e_2*g = g*e_2 = g, \forall g \in G$ $\endgroup$ Commented Sep 21, 2017 at 18:37
  • $\begingroup$ Well I might be misinterpreting it, but I am curious about the case I mentioned, so maybe I should make a new question about it. $\endgroup$ Commented Sep 21, 2017 at 18:45
  • 1
    $\begingroup$ @mathreadler And incidentally, i do think I've seen two operations that interassociate in category theory. I just wish I could remember the keywords... $\endgroup$
    – rschwieb
    Commented Sep 21, 2017 at 18:53

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .