# Tagged Questions

This is the property shared by many binary operations including group operations. For a binary operation $\cdot$, associativity holds if $(x\cdot y)\cdot z = x \cdot(y\cdot z)$.

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### Does associativity imply closure?

Does associativity of binary operation imply closure under this operation? Sometimes definitions of semigroup, group or vector space omit axiom of closure under corresponding operations and sometimes ...
1answer
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### Is there a category theory notion of the image of an axiom or predicate under a functor?

Let me first state that I am a category theory novice so your patience is appreciated. I might be making some very basic conceptional mistakes and I might just need a simpler language to do what I ...
3answers
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### A binary operation, closed over the reals, that is associative, but not commutative

I am aware that matrix multiplication as well as function composition is associative, but not commutative, but are there any other binary operations, specifically that are closed over the reals, that ...
7answers
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### Does associativity imply commutativity?

I used to think that commutativity and associativity are two distinct properties. But recently, I started thinking of something which has troubled this idea: $$(1+1)+1 = 1+ (1+1)\implies 2+1=1+2$$ ...
1answer
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### When a loop with inverse property is commutative

Question How to prove that a loop $L$ with inverse property and $x^3=e$ for all $x$ is commutative iff $(x y)^2=x^2 y^2$ for all $x,y$? Definitions: A loop is a quasigroup with identity $e$. ...
0answers
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### Connected sum $S_1$ # $S_2$ is commutative and associative

The connected sum of two surfaces $S_1$ and $S_2$ is formed by removing a circular hole from each surface and identifying the boundaries together Show that the connected sum $S_1$ # $S_2$ is ...
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### Anti-associativity and a sign problem

I'm doing some algebraic manipulations, and I'm getting crazy over a stupid sign error. I think I've located the source of the problem. It should come from an error I'm making (but can't see) in the ...
0answers
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### Inductive proof of associativity of free groups

I'm really struggling with the inductive proof of the associativity of free groups, given about halfway down page 6 of this pdf. The bit I'm not getting is this: Suppose now that bc involves a ...
6answers
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### Associative Property

What I learn as a basic theory of associative is that $$(a \times b) \times c = a \times (b \times c) \text{ and } (a+b)+c = a+(b+c).$$ However when doing my exercises on this topic, I came across ...
0answers
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### jacobi identity implies flexible algebra

I was looking in the wikipedia article about Non-associative Algebra and came across this interesting line: Each of the properties associative, commutative, anticommutative, Jordan identity, and ...
1answer
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