Mathematical Methods for Physicists: A concise introduction - Site Map

ELEMENTS OF GROUP THEORY
This can be proved as follows. Let G be a group of order n with elements g 1 …ˆE†,
g 2 ; ...; g n : Let H, of order m, be a subgroup of G with elements h 1 …ˆE†,
h 2 ; ...; h m . Now form the set gh k …0 k m†, where g is any element of G not
in H. This collection of elements is called the left-coset of H with respect to g (the
left-coset, because g is at the left of h k ).
If such an element g does not exist, then H ˆ G, and the theorem holds trivially.
If g does exist, than the elements gh k are all di€erent. Otherwise, we would have
gh k ˆ gh`, or h k ˆ h`, which contradicts our assumption that H is a group.
Moreover, the elements gh k are not elements of H. Otherwise, gh k ˆ h j , and we
have
g ˆ h j =h k :
This implies that g is an element of H, which contradicts our assumption that g
does not belong to H.
This left-coset of H does not form a group because it does not contain the
identity element (g 1 ˆ h 1 ˆ E†. If it did form a group, it would require for some h j
such that gh j ˆ E or, equivalently, g ˆ h 1
j . This requires g to be an element of H.
Again this is contrary to assumption that g does not belong to H.
Now every element g in G but not in H belongs to some coset gH. Thus G is a
union of H and a number of non-overlapping cosets, each having m di€erent
elements. The order of G is therefore divisible by m. This proves that the order
of a subgroup is a factor of the order of the full group. The ratio n/m is the index
of H in G.
It is straightforward to prove that a group of order p, where p is a prime
number, has no subgroup. It could be a cyclic group generated by an element a
of period p.
Conjugate classes and invariant subgroups
Another way of dividing a group into subsets is to use the concept of classes. Let
a, b, and u be any three elements of a group, and if
b ˆ u 1 au;
b is said to be the transform of a by the element u; a and b are conjugate (or
equivalent) to each other. It is straightforward to prove that conjugate has the
following three properties:
(1) Every element is conjugate with itself (re¯exivity). Allowing u to be the
identity element E, then we have a ˆ E 1 aE:
(2) If a is conjugate to b, then b is conjugate to a (symmetry). If a ˆ u 1 bu, then
b ˆ uau 1 ˆ…u 1 † 1 a…u 1 †, where u 1 is an element of G if u is.
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