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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:
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 h_{1} and h_{2} belonging to H such that z=xh_{1} and z=yh_{2}. Thus xh_{1}=yh_{2} and hence y = xh_{1}h_{2}^{1}. Now let w be any element of yH. The there exists h_{3} such that w=yh_{3}. But this implies that w=(xh_{1}h_{2}^{1})h_{3}=x(h_{1}h_{2}^{1}h_{3}). Since h_{1}h_{2}^{1}h_{3} 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 h_{1} and h_{2}, z=xh_{1} and z=xh_{2} because this would imply that x=xh_{1}h_{2}^{1} and thus h_{1}h_{2}^{1}=e and hence h_{1}=h_{2}.
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 multiple is the number of cosets of a subgroup.
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