# A question about groups: may I substitute a binary operation with a function?

I have a fundamental question about groups. Consider the definition from Wolfram Mathematica:

A group is a finite or infinite set of elements together with a binary operation (called the group operation) that together satisfy the four fundamental properties of closure, associativity, the identity property, and the inverse property.

In this definition, may I substitute a binary operation with a function, say something like $f(a,b)$ where $f$ is not necessarily a simple operator like addition?

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A binary operation on a set $S$ is a function $f\colon S \times S \to S$. We write $ab$ or $a + b$ (the latter is usually reserved for abelian groups) instead of $f(a, b)$ because it's shorter, but even addition on the integers can be thought of as a function in this way. – Dylan Moreland Jan 7 '12 at 2:27
Is your question really about groups, or are you actually asking 'What can be a binary operation?' – matt Jan 7 '12 at 2:30
My question is about binary operation in the context of group theory. – Ravi Kulkarni Jan 7 '12 at 2:32
Sometimes functions arising in special contexts are given specific names; apart from the vanilla function, one typically comes across terms like operation, operator, map, transform, and transformation. – Srivatsan Jan 7 '12 at 2:35
To be pedantic, the term "algebraic group" generally refers to a particular kind of group, as laid out in the Wikipedia article en.wikipedia.org/wiki/Algebraic_group – Brad Jan 7 '12 at 3:33

A couple of points. First, you write "may I substitute a binary operation with a function". Note that a binary operation is a function, satisfying some extra properties. And there are plenty of contexts where it isn't something as simple as addition. One of the first examples of abstract groups that you meet are permutation groups, where the operation is composition of permutations. This is indeed just one of many examples where the elements of the group are functions of some kind (in this case, a permutation is just a bijection on some finite set). Since function composition is rarely abelian, this is one way that non-abelian structure arises "in nature". If you want to be cute about it, by Cayley's theorem that all finite groups embed into a permutation group, all group laws on finite groups can be thought of as function composition.

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NKS, you nailed it. Thanks everyone. – Ravi Kulkarni Jan 7 '12 at 2:44
In the context of group actions, Cayley's theorum becomes even simpler to state: It says every group is completely determined by the group action of the permutation group of its underlying set onto itself. – Mathemagician1234 Nov 21 '13 at 6:35

I guess the biggest reason for writing $a+b / ab$ is the associativity law looks uncluttered.

Here's function notation for associativity: $f(a,f(b,c)) = f(f(a,b),c)$.

Here's 'normal' notation $(ab)c = a(bc)$ - which one looks better?

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I think the notation $a+b$ et $ab$ are very old, so I'm guessing things happened the other way around : the notation $a+b$ and $ab$ existed for a very long time, and at some point (when came abstract algebra) we realized associativity was what made the 'binary operator' notation more practical that function one. – Joel Cohen Jan 7 '12 at 17:56