Why do composition of two unaries become binary operation?

Question

My textbooks introduces a concept of a 'group'.

As an obvious example, equilateral triangle ABC is used along with all the 6 symmetries it has. Say, I got $f : \{ (A, B); (B, C); (C, A); \}$. It is unary since depends upon exactly 1 argument. Then the composition $f \circ f$ considered: $f \circ f : \{ (A, f[f(A)]); (B, f([f(B)]); (C, f([f(C)]); \}$, which is still unary, for exmaple: $(f \circ f) (A) = f (f (A)) = f(B) = C$. And then my textbooks jumps outright to the group's definition, saying that there should exist a binary operation and whatnot more.

As far as I understand, a binary operation is such $f(x, y)$ that $y$ can not be computed even though one knows $x$'s value. So both variables are independent, like $T \land F$ from the Boolean algebra. This requirement does not hold for the case above.

So, could you help me out with a confusion I described? Maybe it makes sense to consider some other groups like $<\mathbb{Z}, +>$ for clarification?

Answer 1

From your comments on the other answer, it might be less confusing to consider the group $(\mathbb{Z}, +)$. Here the elements of the group are numbers, and addition is clearly a binary operation: you need to add two numbers $x+y$.

In the case of triangle symmetries, the elements of the group are themselves unary operations on the vertices, and composition is a binary operation on unary operations. Much more confusing.

Answer 2

The unaries you are talking about are symmetries: they send a vertex to another vertex. In your example, $f$ sends the vertex $A$ to vertex $B$.

The binary operation your textbook must be talking about (I am only guessing, but I think this is a good guess) is a binary operation that takes two symmetries and returns another symmetry. In your case, the binary operation is function composition. You took the pair $(f, f)$ and then applied function composition to it, yielding a new symmetry $g$ of which you calculated $g(A) = C$.

Answer 3

I think (based especially on your comments to RGS's answer) you're having a bit of a type confusion (EDIT: not really, see comment below):

A group is a pair $(A, *)$ where $A$ is a set - of any type whatsoever - and $*$ is a binary operation on $A$ (satisfying certain properties). One perfectly normal example is:

$A$ is $\mathbb{Z}$ and $*$ is $+$. It's easy to check that the group axioms hold: e.g. $a+(b+c)=(a+b)+c$.

However, the following is also a great example:

$A$ is the set of bijective functions from $\mathbb{Z}$ to $\mathbb{Z}$ and $*$ is composition. (This is similar in spirit to the example you're asking about; I think while this one is a bit more abstract, thinking about it will help build intuition.) Here, even though elements of $A$ are functions, as far as the group is concerned they're just "things." And composition takes two "things" and spits out a third "thing," so in principle this could be a group (and indeed it is).

I think your confusion is that you're expecting elements of the group to always "not look like functions." Part of what makes group theory so powerful, though, is that they can. For example, Galois theory (which was the original motivation for group theory!) shows how we can prove things about fields by looking at groups of functions from fields to fields.