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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.

Doubt
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    For either operation, but $e$ must not be the same for both operations? Maybe we should call them $e_*$ and $e_+$ respectively. – mathreadler Sep 21 '17 at 18:05
  • 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? – Somos Sep 21 '17 at 18:52

1 Answers1

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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 $a,c$.

rschwieb
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  • 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. – mathreadler Sep 21 '17 at 17:58
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    @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. – rschwieb Sep 21 '17 at 18:30
  • 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. $e_2g = ge_2 = g, \forall g \in G$ – mathreadler Sep 21 '17 at 18:37
  • 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. – mathreadler Sep 21 '17 at 18:45
  • @mathreadler Yes, it is strangely worded. To me, though, "have a common identity" seems to mean "common between operations." It could be that this OP is using it to mean "common for all elements" which would be pretty weird. Depending on the OP's clarification, I will leave or remove this answer. – rschwieb Sep 21 '17 at 18:52
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    @mathreadler And incidentally, i do think I've seen two operations that interassociate in category theory. I just wish I could remember the keywords... – rschwieb Sep 21 '17 at 18:53
  • I am so new to algebra so you probably made the right interpretation. – mathreadler Sep 21 '17 at 18:59