# Lagrange's theorem

The set of Lagrange is a mathematical theorem of group theory . In its simplest form it says that the cardinality (or order ) of each subgroup of a finite group shares its cardinality. It was named after the Italian mathematician Joseph-Louis Lagrange .

## statement

There are a group , a subgroup of , the Index of in , so the number of cosets of in , and the group order will be using designated. ${\ displaystyle G}$ ${\ displaystyle H}$ ${\ displaystyle G}$ ${\ displaystyle \ left [G \ colon H \ right]}$ ${\ displaystyle H}$ ${\ displaystyle G}$ ${\ displaystyle H}$ ${\ displaystyle G}$ ${\ displaystyle | H |}$ Then applies

${\ displaystyle \ left | G \ right | = [G: H] \ cdot | H |}$ .

In particular, for both and also divisors of . ${\ displaystyle | G | <\ infty}$ ${\ displaystyle | H |}$ ${\ displaystyle [G: H]}$ ${\ displaystyle | G |}$ ## Proof of the theorem

For each, consider the left minor class . ${\ displaystyle g \ in G}$ ${\ displaystyle gH = \ {gh \ mid h \ in H \}}$ It is a bijection between and , because the mapping is surjective due to the definition of a left secondary class and, according to the truncation rule, also injective. Thus all left secondary classes have the same power as the subgroup . ${\ displaystyle h \ mapsto gh}$ ${\ displaystyle H}$ ${\ displaystyle gH}$ ${\ displaystyle gh_ {1} = gh_ {2} \ Rightarrow h_ {1} = h_ {2}}$ ${\ displaystyle | H |}$ Since the secondary classes can be defined as equivalence classes of the equivalence relation , they provide a partition of . If one chooses a representative system of the secondary classes with the help of the axiom of choice , then one has a bijection between and through the mapping . After the definition of the index and the system of representatives, the following applies and thus one obtains ${\ displaystyle a \ sim b \ Leftrightarrow a ^ {- 1} b \ in H}$ ${\ displaystyle G}$ ${\ displaystyle R}$ ${\ displaystyle R \ times H}$ ${\ displaystyle G}$ ${\ displaystyle (r, h) \ mapsto rh}$ ${\ displaystyle \ left [G: H \ right] = \ left | R \ right |}$ ${\ displaystyle \ left | G \ right | = \ left | R \ times H \ right | = \ left | R \ right | \ cdot \ left | H \ right | = \ left [G: H \ right] \ cdot \ left | H \ right |}$ which was to be proved.

## Inferences

Since the order of a group element is precisely the order of the subgroup that is generated by this element, it follows from Lagrange's theorem that the order of a group element always shares the order of the group.

From this result one gets directly the little Fermat's theorem from number theory and as a further generalization Euler's theorem .

Finite groups, whose group order is a prime number , are cyclic and simple according to Lagrange's theorem . Since the group order is a prime number, according to Lagrange's theorem, there can only be the trivial subgroups and thus every non-neutral element already creates the whole group and there are only the trivial normal divisors.

## generalization

Be a group, subgroups. Then one obtains by applying Lagrange's theorem twice ${\ displaystyle G}$ ${\ displaystyle U \ leq V \ leq G}$ ${\ displaystyle \ left [G: U \ right] = \ left [G: V \ right] \ cdot \ left [V: U \ right]}$ If you choose , you get Lagrange's theorem again. ${\ displaystyle U = \ left \ {e \ right \}}$ ## Subgroups of a given order

With Lagrange's theorem one has a necessary criterion for the existence of a subgroup of a certain order for finite groups . However, the criterion is not sufficient , that is, in general, for finite groups there is not a subgroup that has this order for every divisor of the group order. The smallest group that makes this clear is the group . has elements, but no subgroup of order . ${\ displaystyle A_ {4}}$ ${\ displaystyle A_ {4}}$ ${\ displaystyle 12}$ ${\ displaystyle 6}$ Nevertheless, there are certain groups, which for every part of the group order also have a subgroup of this order. The cyclic groups are an example . There are also sentences that guarantee the existence of subgroups of certain orders. An example of this are the Sylow sentences .