7 Groups 57We now show that a nonemptyfinitesubsetHof a groupGis a subgroup ofGif
it is closed under multiplication only. For, ifa∈H,thenha∈Hfor allh∈H.Since
His finite and the mappingh→haofHinto itself is injective, it is also surjective by
the pigeonhole principle (Corollary I.6). Henceha=afor someh∈H,whichshows
thatHcontains the identity element ofG. It now further follows thatha=efor some
h∈H, which shows thatHis also closed under inversion.
A group is said to befiniteif it contains only finitely many elements and to be of
order nif it contains exactlynelements.
In order to give an important example of a subgroup we digress briefly. Letnbe
a positive integer and letAbe the set{ 1 , 2 ,...,n}with the elements in their natural
order. Since we regardAas ordered, a bijective mapα:A→Awill be called aper-
mutation. The set of all permutations ofAis a group under composition, thesymmetric
groupSn.
Suppose now thatn>1. An inversion of order induced by the permutationαis a
pair (i,j) withi<jfor whichα(i)>α(j). The permutationαis said to beevenor
oddaccording as the total number of inversions of order is even or odd. For example,
the permutation{ 1 , 2 , 3 , 4 , 5 }→{ 3 , 5 , 4 , 1 , 2 }is odd, since there are 2+ 3 + 2 = 7
inversions of order.
Thesignof the permutationαis defined by
sgn(α)=1or−1 according asαis even or odd.Evidently we can write
sgn(α)=∏
1 ≤i<j≤n{α(j)−α(i)}/(j−i),from which it follows that
sgn(αβ)=sgn(α)sgn(β).Since the sign of the identity permutation is 1, this implies
sgn(α−^1 )=sgn(α).Thus sgn(ρ−^1 αρ)=sgn(α)for any permutationρofA, and so sgn(α)is actually
independent of the ordering ofA.
Since the product of two even permutations is again an even permutation, the even
permutations form a subgroup ofSn,thealternating groupAn. The order ofAnis
n!/2. For letτbe the permutation{ 1 , 2 , 3 ,...,n}→{ 2 , 1 , 3 ,...,n}. Since there is
only one inversion of order,τis odd. Sinceττis the identity permutation, a permuta-
tion is odd if and only if it has the formατ,whereαis even. Hence the number of odd
permutations is equal to the number of even permutations.
It may be mentioned that the sign of a permutation can also be determined without
actually counting the total number of inversions. In fact anyα∈Snmay be written as
a product ofvdisjoint cycles, andαis even or odd according asn−vis even or odd.
We now return to the main story. LetHbe a subgroup of an arbitrary groupGand
leta,bbe elements ofG. We writea∼rbifba−^1 ∈H. We will show that this is an
equivalence relation.