I am learning about Groups from a discrete mathematics textbook for a computer science course by Grimaldi.

An algebraic system is defined as a set along with operations on elements of the set. An algebraic structure is defined as a set, operations on elements of the set and relations between elements of the set which leads to a structure on the elements of a set.

The author then goes on to talk about groups as being algebraic structures(so does wikipedia). I am failing to see what structure a group imposes on a set it is defined on. This is making me think about a group an algebraic system instead of an algebraic structure.

What should I see in a group that will help me relate it the definition of an algebraic structure? What should I understand by "structure on the elements of a set"? Explanation with examples would help a lot.

EDIT - Exact definitions as given in the textbook
An algebraic system is a system consisting of a nonempty set $A$ and one or more n-ary operations on the set $A$. It is denoted by $ \langle A, f_1, f_2, ... \rangle$.
An algebraic structure is an algebraic system, $\langle A, f_1, f_2, ... , R_1, R_2, ...\rangle$, wherein addition to operations $f_i$, the relations $R_i$ are defined on A. This leads to a structure on the elements of A.

  • $\begingroup$ Your definitions are rather vague. Can you give exact quotations? $\endgroup$ – Zhen Lin Dec 19 '12 at 5:13
  • $\begingroup$ @ZhenLin, I've added the exact definitions. $\endgroup$ – Abhijith Madhav Dec 19 '12 at 5:45
  • 1
    $\begingroup$ Hmmm. I think by ‘algebraic structure’ most people mean a set with operations but no relations. $\endgroup$ – Zhen Lin Dec 19 '12 at 7:33
  • $\begingroup$ Following the definitions, an algebraic system is for example $\langle \Bbb Z, +\rangle$ and an algebraic structure is $\langle \Bbb Z,+,\leq\rangle$ $\endgroup$ – leo Dec 19 '12 at 20:34

I am not really familiar with the 'algebraic system' terminology, but it seems as though the difference between that and an 'algebraic structure' is that an algebraic system can be a set with any binary operation on that set, whereas the operation on a set in an algebraic structure must conform to certain rules (in the case of groups, inverses, associativity, and identity).

Maybe the easiest example is the integers. As a set, $\mathbb{Z}$ is just a collection of elements which have nothing to do with each other - $0$ is no more special than $79$, $10$ and $-10$ have no particular relationship. But once we consider the integers as a group under the binary operation 'addition' (which I think we're all familiar with) we are able to see that $10+(-10)=0$ and $z + 0 = z$ for every $z\in \mathbb{Z}$.

  • $\begingroup$ You probably mean addition modulo $n$ in your definition of $\mathbb Z_n$. Also, all groups can be realized as groups of permutations, not just finite groups. $\endgroup$ – Santiago Canez Dec 19 '12 at 5:01
  • $\begingroup$ @AlexanderGruber, So is it correct of me to say that a set of an algebraic structure is different from an ordinary set in that its axioms differentiate certain elements(identity and inverses in groups) and sets rules about how operators behave(closure and associativity in groups)? This differentiation and behaviour specification is what is alluded to by "structure"? $\endgroup$ – Abhijith Madhav Dec 19 '12 at 5:21
  • $\begingroup$ More or less. If we're getting technical it is the operation which imposes the relation on the set, but I think you have the concept. $\endgroup$ – Alexander Gruber Dec 19 '12 at 5:23
  • $\begingroup$ @AlexanderGruber, Thanks. I had recently learned about relations and partial orders in particular and was subconsciously expecting the 'structure' to do something similar to imposing an order on the elements of the set. I could not visualize 'structure' as something other than an order until I read the third paragraph of your answer. My question seems so silly now. Also would you consider maybe editing your answer to remove the 1st, 4th and 5th paragraphs before I accept it. I felt that they were not directly relevant to my question. $\endgroup$ – Abhijith Madhav Dec 19 '12 at 5:32
  • $\begingroup$ @Abhijith Whatever you say boss. $\endgroup$ – Alexander Gruber Dec 19 '12 at 5:40

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