Example

In any monoid E the set of invertible elements with the structure induced by that on E is a group. In particular, the set of bijective mappings of a set F onto itself (or set of permutations of F) is a group under the law $(f, g) \mapsto f \circ g$, called the symmetric group of the set F and denoted by $\mathfrak{S}_F$.

In this paragraph, unless otherwise indicated, the law of composition of a group will always be written multiplicatively and $e$ will denote the identity element of such a group law.

A group G is called finite if the underlying set of G is finite; otherwise it is called infinite; the cardinal of a group is called the order of the group.

If a law of composition on G determines a group structure on G, so does the opposite law. The mapping of a group G onto itself which associates with each $x \in G$ the inverse of $x$ is an isomorphism of G onto the opposite group (§ 2, no. 3, Proposition 4).

Following our general conventions (Set Theory, II, § 3, no. 1), we shall denote by $A^{-1}$ the image of a subset A of G under the mapping $x \mapsto x^{-1}$. But it is important to note that, in spite of the analogy of notation, $A^{-1}$ is definitely not the inverse element of A under the law of composition $(X, Y) \mapsto XY$ between subsets of G (recall that XY is the set of $xy$ with $x \in X, y \in Y$): the identity element under this law is $\{e\}$ and the only invertible elements of $\mathscr{P}(G)$ under this law are the sets A consisting of a single element (such an A, moreover, certainly has inverse $A^{-1}$). The identity

$(AB)^{-1}=B^{-1}A^{-1}$ holds for $A\subset G$, $B\subset G$. $A$ is called a symmetric subset of $G$ if $A=A^{-1}$. For all $A\subset G$, $A\cup A^{-1}$, $A\cap A^{-1}$ and $AA^{-1}$ are symmetric.