Mathematical Methods for Physics and Engineering : A Comprehensive Guide

28.7 SUBDIVIDING A GROUP

• Two cosetsX 1 HandX 2 Hare identical if, and only if,X 2 −^1 X 1 belongs toH.If

X 2 −^1 X 1 belongs toHthenX 1 =X 2 Yifor somei,and

X 1 H=X 2 YiH=X 2 H,

since by the permutation lawYiH=H. Thus the two cosets are identical.

Conversely, supposeX 1 H=X 2 H.ThenX− 21 X 1 H=H. But one element of

H(on the left of the equation) isI; thusX− 21 X 1 must also be an element ofH

(on the right). This proves the stated result.

• Every element ofGis in some left cosetXH.This follows trivially sinceH

containsI, and so the elementXiis in the cosetXiH.

The final step in establishing Lagrange’s theorem is, as previously, to note that

each coset containshelements, that the cosets are disjoint and that every one of

thegelements inGappears in one and only one distinct coset. It follows that

g=khfor some integerk.

As noted earlier, Lagrange’s theorem justifies our statement that any group of

orderp,wherepis prime, must be cyclic and cannot have any proper subgroups:

since any subgroup must have an order that dividesp, this can only be 1 orp,

corresponding to the two trivial subgroupsIand the whole group.

It may be helpful to see an example worked through explicitly, and we again

use the same six-element group.

Find the left cosets of the proper subgroupHof the groupGthat has table 28.8 as its

multiplication table.

The subgroup consists of the set of elementsH={I, A,B}. We note in passing that it has
order 3, which, as required by Lagrange’s theorem, is a divisor of 6, the order ofG.Asin
all cases,Hitself provides the first (left) coset, formally the coset

IH={II,IA,IB}={I, A,B}.

We continue by choosing an element not already selected,Csay, and form

CH={CI,CA, CB}={C, D, E}.

These two cosets ofHexhaustG, and are therefore the only cosets, the index ofHinG
being equal to 2.
This completes the example, but it is useful to demonstrate that it would not have
mattered if we had takenD, say, instead ofIto form a first coset

DH={DI, DA, DB}={D, E, C},

and then, from previously unselected elements, pickedB, say:

BH={BI, BA, BB}={B, I, A}.

The same two cosets would have resulted.

It will be noticed that the cosets are the same groupings of the elements

ofGwhich we earlier noted as being the choice of adjacent column and row

headings that give the multiplication table its ‘neatest’ appearance. Furthermore,