Are integers relevant for every Group?

Question

The definition of the order of an element in a group is:

The order of an element $x$ of a group $G$ is the smallest positive integer $n$ such that $x^{n}=e$.

Doesn't this definition assume that the integers are somehow relevant to every group?

All of the other definitions concerning Groups do not invoke the integers in any way.

I would expect that Groups would contain elements that would somehow represent exponentiation.

This question may reflect some deep misunderstandings on my part so please feel free to offer guidance not directly relevant to the question.

Answer 1

Every group has an associative "multiplication" (binary operation), so there is no ambiguity in referring to the integer power of an element.

In particular, $x^n$ is just short hand for $$ x^n = \overbrace{x x \cdots x}^n $$ where we note that this expression is this same, no matter how the $x$s are "grouped together" within that product. This only defines exponentiation for positive integers; however, because groups have an identity and group elements have inverses, we may also define $x^0$ to be the identity element and $$ x^{-n} = (x^{-1})^{n} $$ for any $n \geq 1$.

Answer 2

Here is an alternative viewpoint:

Actually, the common definition of a group with an operation $\cdot : G\times G\to G$ is not the only possible definition of a group. In fact it is biased, because for any group there a alot of $n$-ary operations besides the group multiplication. For example you have the following operations:

• $^{-1} : G\to G, x\mapsto x^{-1}$
• $/ : G^2 \to G, (x,y)\mapsto xy^{-1}$
• $G^2\to G, (x,y) \mapsto xyx^{-1}$
• $G^3 \to G, (x,y,z) \mapsto xyz^{-1}$
• $G^4 \to G, (x_1,x_2,x_3,x_4) \mapsto x_{1} x_{2} x_{3} x_{4}$
• ...

but there is no end to this list! Furthermore the multiplication $\cdot$ is not special among all these operations. You can state the group axioms with a lot of these other operations as well (division $/$ is a prime candidate)

It begs the question: is there an "ultimate" operation that stands above all of the other? The answer is: Yes! (sort of):

Let $G^* = \{ (x_1,\dots, x_n) : x_1,\dots, x_n\in G\}$. That is the set of $n$-tuples or words over $G$. Then there is an operation:

$$\prod : G^* \to G, (x_1,\dots, x_n) \mapsto x_1\cdots x_n$$ that maps a word of symbols, that is a bunch of group elements, to their multiplication. (Yes, it is kind of cheating because I still need the group multiplication, but maybe you get the idea). We commonly write: $$\prod_{i=1}^n x_i\text{ for }\prod (x_1,\dots, x_n)$$

The associativity says, that $\prod$ is well-defined (bracketing does not matter). The identity is also encoded in $\prod$! We have: $$1_G := \prod ()$$ where $()$ is the empty word (the tuple of length $0$). This is a convenient definition making the recursive formula work: $$\prod_{i=1}^n x_i = x_1 \cdot \prod_{i=2}^n x_i$$ even when $n = 1$, because: $$x_1\cdot \prod_{i=2}^1 x_i = x_1\cdot \prod () = x_1\cdot 1_G = x_1 = \prod_{i=1}^1 x_i$$ Note, that $x^n = \prod_{i=1}^n x$.

This should make obvious how nonnegative integers come into play (as lengths of words that I can put into $\prod$)

(Even if you do not believe all of this is useful yet, you will sort of rediscover all of this when you take a look at free groups.)