Section 1.6. Quotient Groups 57 For the reverse inclusion, if f([c]) = ([c]m, [c]n) U(Im) × U(In), then we must show that [c] U(Imn). There is [d]m Im with [c]m[d]m = [1]m, and there is [e]n In with [c]n[e]n = [1]n. Since f is surjective, there is b Z with ([b]m, [b]n) = ([d]m, [e]n), so that f([1]) = ([1]m, [1]n) = ([c]m[b]m, [c]n[b]n) = f([c][b]). Since f is an injection, [1] = [c][b] and [c] U(Imn). Corollary 1.91. (i) If p is prime, then φ(pe) = pe pe−1 = pe 1 1 p . (ii) If n = p11 e · · · ptt e is the prime factorization, where p1,...,pt are distinct primes, then φ(n) = n 1 1 p1 · · · 1 1 pt . Proof. Part (i) holds because (k, pe) = 1 if and only if p k, while part (ii) follows from Corollary 1.90. Lemma 1.92. Let G = a be a cyclic group. (i) Every subgroup S of G is cyclic. (ii) If |G| = n, then G has a unique subgroup of order d for each divisor d of n. Proof. (i) We may assume that S = {1}. Each element s S, as every element of G, is a power of a. If m is the smallest positive integer with am S, we claim that S = am . Clearly, am S. For the reverse inclusion, let s = ak S. By the Division Algorithm, k = qm + r, where 0 r m. Hence, s = ak = amqar = ar. If r 0, we contradict the minimality of m. Thus, k = qm and s = ak = (am)q am . (ii) If n = cd, we show that ac has order d (whence ac is a subgroup of order d). Clearly (ac)d = acd = an = 1 we claim that d is the smallest such power. If (ac)m = 1, where m d, then n | cm, by Proposition 1.26 hence cm = ns = dcs for some integer s, and m = ds d, a contradiction. To prove uniqueness, assume that x is a subgroup of order d [every subgroup is cyclic, by part (i)]. Now x = am and 1 = xd = amd hence md = nk for some integer k. Therefore, x = am = (an/d)k = (ac)k, so that x ac . Since both subgroups have the same order d, it follows that x = ac . The next theorem will be used to prove Theorem 2.46: the multiplicative group Ip × is cyclic if p is prime. We will use Proposition 2.71(iii) in the next proof it says that n = d|n φ(d) for every integer n 1. Theorem 1.93. A group G of order n is cyclic if and only if, for each divisor d of n, there is at most one cyclic subgroup of order d. Proof. If G is cyclic, then the result follows from Lemma 1.92.
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