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The Lagrange Theorem for Groups

The order of a group is the number of elements in the group. A group may have subgroups within it. The orders of the subgroups of a group cannot be just any values. The order of a subgroup must be a factor of the order of the group. That is to say, a group of order 8 cannot have a subgroup of order 5 but it could have subgroups of orders 2 and 4. This is the essense of the Lagrange Theorem for groups:

The order of a subgroup of a group must be
a factor of the order of the group.

Proof: Let H be a subgroup of a group G. A left coset of H for an element g of G is the set gH={gh: h∈H}. It needs to be shown that the left cosets constitute a partitioning of the set G; i.e., each element of G belongs to one and only one coset.

First, any element g must belong to some coset. Obviously g belongs to its own coset because g=ge, where e is the identity element of H.

Consider two left cosets xH and yH. Suppose xH and yH have a common element z. This means that there exists h1 and h2 belonging to H such that z=xh1 and z=yh2. Thus xh1=yh2 and hence y = xh1h2-1. Now let w be any element of yH. The there exists h3 such that w=yh3. But this implies that w=(xh1h2-1)h3=x(h1h2-1h3). Since h1h2-1h3 necessarily belongs to H w must belong to xH. Thus yH ⊆ xH. Likewise a similar argument demonstrates that xH ⊆ yH so xH=yH.

The cardinality of all the cosets must be equal. It cannot be that there is an element z of xH such that for two different h1 and h2, z=xh1 and z=xh2 because this would imply that x=xh1h2-1 and thus h1h2-1=e and hence h1=h2.

Thus G is partitioned into cosets of equal size (which is the order of the subgroup) and hence the order of G must be a multiple of the order of the subgroup.

Since the orders of a group and a subgroup may be infinite, it is best to state the theorem in the form

The order of a group must be a multiple
of the order of any of its subgroups.

The multiple is the number of cosets of a subgroup.

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