Understanding integers as a group in light of a general definition

Question

The place from where I am studying Abstract Algebra recommends the following as the most general understanding of a group, the rest reducible to this.

For a set $T$, the set of all bijections $Sym(T)$ from $T$ to $T$ constitutes a group under the product operation being the composition of these bijections. Identity is then a bijection onto itself.

But when thinking of a group in terms of bijections or transformations, how do we reconcile this understanding to understand the Integers being a group under addition? What is the bijection here? Am I missing something here?

Answer 1

There is a correspondence between integers and a certain class of bijections of $\mathbb{Z}$, as follows. For $n \in \mathbb{Z}$, define $f_n:\mathbb{Z} \to \mathbb{Z}$ by $f_n(k) = k+n$. Then $f_n$ is a bijection. Geometrially, $f_n$ translates $\mathbb{Z}$ to the right by $n$. Then the association $n \mapsto f_n$ is a way to think of each integer as a bijection $\mathbb{Z} \to \mathbb{Z}$.

Answer 2

That definition is either wrong (as written) or rather misleading. Here is the usual definition:

A group is a set $G$ with a law of composition $\cdot: G^2 \to G$ such that

• Identity: there is an element $1\in G$ which satisfies $g\cdot 1 = 1\cdot g=g$ for all $g\in G$
• Inverse: for every $g\in G$ there is an element $h\in G$ such that $g\cdot h=h\cdot g = 1$ (this element is unique and is usually written $g^{-1}$)

It is sometimes convenient to write a group as a quadruple $(G,\cdot,-^{-1},1)$.

An example of a group is the group of bijections $\mathrm{Sym}(T)$ on a set $T$, with the law of composition given by function composition. The identity is exactly the identity map $\mathrm{id}_T$ and the inverse of a bijection $f: T \to T$ is exactly the inverse map $f^{-1}: T \to T$.

There are certainly groups which are not of the above form. One way of seeing this is that for $|T|>2$ the group $\mathrm{Sym}(T)$ is not Abelian, meaning that the law of composition is not commutative. This is because given distinct elements $x,y,z\in T$, we can define bijections $f_1,f_2: T\to T$ by $$f_1(x)=y,f_1(y)=x, f_1(w)=w \quad \text{for $w\in T\setminus\{x,y\}$}$$ $$f_1(y)=z,f_1(z)=y, f_1(w)=w \quad \text{for $w\in T\setminus\{y,z\}$}$$ In other words, $f_1$ switches $x$ and $y$ and leaves everything else fixed, and $f_2$ switches $y$ and $z$ and leaves everything else fixed. Then $$(f_1\circ f_2)(y) = f_1(f_2(y))=f_1(z)=z$$ $$(f_2\circ f_1)(y) = f_2(f_1(y))=f_2(x)=x$$ are unequal, so $f_1\circ f_2 \neq f_2 \circ f_1$.

However, $\mathbb{Z}$ with addition is Abelian, so it can't be of the form $\mathrm{Sym}(T)$ for any set $T$.

What is true, however, is that every group is (isomorphic) to a subgroup of $\mathrm{Sym}(T)$ for some set $T$. This is known as Cayley's Theorem.

By subgroup of a group $(G,\cdot,-^{-1},1)$ we mean a subset $H\subset G$ for which $H$ is closed under $\cdot$ and taking inverses, and contains $1$.

Given a group $(G,\cdot,-^{-1},1)$, consider $\mathrm{Sym}(G)$. We then associated each element $g\in G$ to the function $\tau_g: G \to G$ defined by $\tau_g(h) = g\cdot h$. This is invertible, since $\tau_g \circ \tau_{g^{-1}} = \mathrm{id}_G$. More generally, $\tau_g \circ \tau_h = \tau_{g\cdot h}$. This allows us to treat $G$ like a subgroup of $\mathrm{Sym}(G)$.

Answer 3

The idea is not to think of a group in terms of bijections. It so happens that the elements of $\mathrm{Sym}(T)$ are bijections. The elements of $\mathbb Z$ are integers. Although there is a relationship between the two, the source you are studying from probably mentions $\mathrm{Sym}(T)$ as a good general exemple of group. In which case, the focus is on the group structure and properties, whatever the nature of its elements is.