Lagrange's theorem

In mathematics, most commonly, Lagrange's theorem states that if G is a finite group and H is a subgroup of G, then the order (that is, the number of elements) of H divides the order of G.

Related Topics:
Mathematics - Group - Subgroup - Order - Divides

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This can be shown using the concept of left cosets of H in G. The left cosets are the equivalence classes of a certain equivalence relation on G and therefore form a partition of G. If we can show that all cosets of H have the same number of elements, then we are done, since H itself is a coset of H. Now, if aH and bH are two left cosets of H, we can define a map f : aH → bH by setting f(x) = ba-1x. This map is bijective because its inverse is given by f -1(y) = ab-1y.

Related Topics:
Coset - Equivalence class - Equivalence relation - Bijective

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This proof also shows that the quotient of the orders |G| / |H| is equal to the index (the number of left cosets of H in G). If we write this statement as

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:|G| = · |H|,

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then, interpreted as a statement about cardinal numbers, it remains true even for infinite groups G and H.

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A consequence of the theorem is that the order of any element a of a finite group (i.e. the smallest positive integer k with ak = e) divides the order of that group, since the order of a is equal to the order of the cyclic subgroup generated by a. If the group has n elements, it follows

Related Topics:
Order of any element - Cyclic - Generated

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:an = e.

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This can be used to prove Fermat's little theorem and its generalization, Euler's theorem.

Related Topics:
Fermat's little theorem - Euler's theorem

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The converse of Lagrange's theorem is not true in general: given a finite group G and a divisor d of |G|, there does not necessarily exist a subgroup of G with order d. The smallest example is the alternating group G = A4 which has 12 elements but no subgroup of order 6. However, if G is abelian, then there always exists a subgroup of order d.

Related Topics:
Alternating group - Abelian

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