# Can a group be defined in terms of a relation on a set?

Wikipedia defines a group as "an algebraic structure consisting of a set together with an operation that combines any two of its elements to form a third element." I keep thinking that there is a connection to this definition and a relation on a set, but I'm not sure what it is. Obviously, relations and operators are connected. Can groups be defined in terms of sets and relations? I am new to this, and the Wikipedia article is over my head.

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Well, everything is connected. I'm not really sure what you're looking for. – Qiaochu Yuan Sep 17 '11 at 1:19
@Qiaochu Yuan In other words, is it possible to define a group as "an algebraic structure consisting of a set together with a relation..."? – Kazark Sep 17 '11 at 1:24
This seems to be relevant (see the answer): math.stackexchange.com/questions/4009/…. – Srivatsan Sep 17 '11 at 1:24
@Srivatsan Narayanan Yes, that is relevant, thank you. So it would seem that the answer to my question is no, relations are more general than operators, and the question becomes, is there there a name for "an algebraic structure consisting of a set together with a relation..."? – Kazark Sep 17 '11 at 1:32
You should look into a branch of mathematical logic called model theory. Within some logical language (often work in first order logic), a set (called a domain) together with relations, and special relations called constants and functions form something called a structure. – William Sep 17 '11 at 1:35

Note that a relation $R$ on a set $G$ is any subset of $G \times G$. A function is a relation on $G \times G$ such that if $(a,b) \in f$ and $(a,c) \in f$, then $b = c$.
Therefore a group is a set $G$ with a relation $*$ which happens to be a function. Moreover, this function satisfies some properties like associativity, etc. Also as is typical in model theory, you often say a group is a set, with a binary function $*$, and a constant $e$, which represents the identity. Again, the constant can still be thought of as a unary relation. You can also define a group to include a symbol for taking an inverse. This can still be thought of as a relation since it is a unary function.
Be careful: a binary relation on $G \times G$ would be a subset of $G \times G \times G \times G$. A binary operation is a ternary relation, i.e. a subset of $G \times G \times G$. – Zhen Lin Sep 17 '11 at 1:55