For a *motion graphic* tutorial on groups in the theory of symmetry see Physics As Symmetry. There group members are 'altered scrutinies' of inspection.

A 'group' is a collection of members which satisfy some prescribed elemental mathematical criteria. (See Exhibit 9 in Physics As Symmetry for a list of the five criteria.) The essential effect of these criteria is that a *multiplication table* (Exhibit 7, Physics As Symmetry) may always be constructed from the members of a group. A *subgroup* consists of a collection of members within the group that, themselves, make up a group. The members, S_{i}, of the subgroup have a complete multiplication table of their own. That S_{i}S_{j} = S_{k} for any i,j is the key property that marks the collection {S} as a subgroup of {G}.

The whole group, G, is parceled into subgroup members S, and non-subgroup ones N. The square brackets [ ] refer to the number of members in a collection. Thus the 'order' or size of {S} is [S]. The 'order' of {G} - the number of its members - is [G].

An important theorem due to Lagrange relates the size, [S], of the subgroup to the size, [G], of the group:

[G] = integer x [S]

What follows is a proof of this theorem.

The ordering of G first by members of S followed by members of N exhibits the subgroup S in the Cayley Table (multiplication table, sequential action table) as a block within the whole table. One corner of the table contains elements belonging only to this subgroup as seen in the figure.

Such collections appear in the multiplication table as the rows in the rectangle array shown in the figure. The significant feature of these collections is this:

*Any two of them have either all elements in common or none in common.*

The statement that "two left cosets have all their members in common or none in common" means that "the collection N_{a}{S} is either equal to the collection N_{b}{S} or disjoint from it".

In the figure N_{2}{S} and N_{c}{S} are disjoint. They have no members in common. But N_{1}{S} and N_{2}{S} have all their members in common.

- Proof:
- Suppose N
_{a}S_{i}= N_{b}S_{j}is one element in common. - then N
_{b}^{-1}N_{a}= S_{j}S_{i}^{-1}= S_{k}so N_{b}^{-1}N_{a}is an element of S - hence N
_{b}^{-1}N_{a}{S} = {S} (since S_{i}{S} = {S} any i) - therefore N
_{a}{S} = N_{b}{S}. All elements are in common

But, since each individual coset has [S] members, the number of elements in this non-repeating coset collection of [N] members is an integer x [S].

Using the model, here's the idea displayed visually. The [S] members of the first coset begin a,b,c ... Write these down. Omit the [S] members b,c,a... and also the [S] members c,a,b... because these cosets contain elements that have already appeared. Include f,h,d... but not h,d,f... for the same reason. In the end you simply have the [N] members a,b,c...f,h,d... But by the construction we have written an integral number of [S] members because each coset written down has [S] members.

([S] members),([S] members), . . .([S] members) = [N] members

So an integer x [S] = [N] and [G] = (that integer +1) x [S] = integer x [S].

The result is expressed geometrically in the diagram below!

August 1997

Marvin Chester

email: chester@physics.ucla.edu

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© m chester 1997 Occidental CA