Question: I was wondering about the ways we say something is a group. For example, the set $$G = \{z \in \mathbb{C} \mid z^n = 1 \text{ for some }n \in \mathbb{Z}^+\}$$ is a group under multiplication. But if I were to just say "Assume that $G = \{1,a,b,c\}$ is a group of order $4$," how can we not specify the binary operation? Is this just saying there exists a binary operation for which it is a group? 

**Edit:** Here is a definition of a group that seems to show what I am saying:

[![enter image description here](https://i.stack.imgur.com/oiB4T.png)](https://i.stack.imgur.com/oiB4T.png)

Answer: 
Yes. If you are assuming $G$ is a group, then it certainly must have the properties of a group by assumption. So, in particular, you are indeed assuming that $G$ must have such a binary operation, even if you do not explicitly state what it is.
"What is a group? Algebraists teach that this is supposedly a set with two operations that satisfy a load of easily-forgettable axioms. This definition provokes a natural protest: why would any sensible person need such pairs of operations? "Oh, curse this maths" - concludes the student (who, possibly, becomes the Minister for Science in the future).

We get a totally different situation if we start off not with the group but with the concept of a transformation (a one-to-one mapping of a set onto itself) as it was historically. A collection of transformations of a set is called a group if along with any two transformations it contains the result of their consecutive application and an inverse transformation along with every transformation.

This is all the definition there is. The so-called "axioms" are in fact just (obvious) properties of groups of transformations."

From V. Arnol'd's article "On teaching mathematics", <http://pauli.uni-muenster.de/~munsteg/arnold.html>