54 Chapter 1. Groups I check that sT π(s)T is such a bijection.28 If T S, then T/K S/K and (S/K)/(T/K) = S/T , by the Third Isomorphism Theorem that is, S∗/T = S/T . It remains to show that if T S∗, then T S that is, if t T and s S, then sts−1 T . Now π(sts−1) = π(s)π(t)π(s)−1 π(s)T ∗π(s)−1 = T ∗, so that sts−1 π−1(T ∗) = T . Example 1.83. Let G = a be a (multiplicative) cyclic group of order 30. If π : Z G is defined by π(n) = an, then ker π = 30 . The subgroups 30 10 2 Z correspond to the subgroups {1} = a30 a10 a2 a . Moreover, the quotient groups are a10 a30 = 10 30 = I3, a2 a10 = 2 10 = I5, a a2 = Z 2 = I2. Here are some applications of the Isomorphism Theorems. Proposition 1.84. If G is a finite abelian group and d is a divisor of |G|, then G contains a subgroup of order d. Remark. We have already seen, in Proposition 1.70, that this proposition can be false for nonabelian groups. Proof. We first prove the result, by induction on |G|, for prime divisors p of |G|. The base step |G| = 1 is true, for there are no prime divisors of 1. For the inductive step, choose a G of order k 1. If p | k, say k = p, then Exercise 1.32 on page 27 says that a has order p. If p k, consider the cyclic subgroup H = a . Now H G, because G is abelian, and so the quotient group G/H exists. Note that |G/H| = |G|/k is divisible by p, and so the inductive hypothesis gives an element bH G/H of order p. If b has order m, then Exercise 1.54 on page 45 gives p | m. We have returned to the first case. Let d be any divisor of |G|, and let p be a prime divisor of d. We have just seen that there is a subgroup S G of order p. Now S G, because G is abelian, and G/S is a group of order n/p. By induction on |G|, G/S has a subgroup H∗ of order d/p. The Correspondence Theorem gives H∗ = H/S for some subgroup H of G containing S, and |H| = |H∗||S| = d. We now construct a new group from two given groups. Definition. If H and K are groups, then their direct product, denoted by H ×K, is the set of all ordered pairs (h, k), with h H and k K, equipped with the operation (h, k)(h , k ) = (hh , kk ). 28When G is finite, we may prove that [S : T ] = [S∗ : T ∗] as follows: [S∗ : T ∗] = |S∗|/|T ∗| = |S/K|/|T/K| = (|S|/|K|) / (|T |/|K|) = |S|/|T | = [S : T ].
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