Suppose that G is a finite group and H is a subgroup of G Pr

Suppose that G is a finite group and H is a subgroup of G. Prove: The order of H divides the order of G. Prove: If G is finite group of prime order then G is cyclic.

Solution

Lagrange\'s theorem, in group theory, states that for any finite group G, the order (number of elements) of every subgroup H of G divides the order of G.

Let H G, let |G| = n, and let |H| =m.

Since every coset (left or right) of a subgroup H G has the same number of elements as H, we know that every coset of H also has m elements. Let r be the number of cells in the partition of G into left cosets of H (because the union of the left cosets of H in G is G, and these cosets are disjoint, they partition G).Then n = r m: i.e., |G| = (G:H)|H| so m=|H| is a divisor of n=|G|.

Let p be a prime and G be a group such that |G| = p. Then G contains more than one element. Let g G such that g eG. Then g contains more than one element. Since g G, by Lagrange’s theorem, |g| divides p. Since |g| >1 and |g| divides a prime, |g| =p = |G|. Hence, g = G. It follows that G is a cyclic group.