Posts Tagged ‘Burnside normal complement theorem’

Groups of order n with gcd(n, phi(n))=1 are cyclic

Posted: December 13, 2010 in Basic Algebra, Groups
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Let \phi be the Euler totient function. Before getting into the main problem we give a useful lemma.

Lemma. Let H be a subgroup of a group G. Let N(H) and C(H) be the normalizer and the centralizer of H in G, respectively. Then C(H) \subseteq N(H) and N(H)/C(H) is isomorphic to a subgroup of \text{Aut}(H).

Proof. Define f: N(H) \longrightarrow \text{Aut}(H) by f(x)(h)=xhx^{-1} for all x \in N(H) and h \in H. See that f is a well-defined group homomorphism and \ker f = C(H). \ \Box

Problem. Let G be a group of order n. Prove that if \gcd(n, \phi(n))=1, then G is cyclic.

Solution (Y.Sharifi). The proof is by induction on n. The case n = 1 is trivial. For n > 1, since \gcd(n, \phi(n))=1, we must have n=p_1 p_2 \cdots p_k for some distinct primes p_i. Let P be a Sylow p_1-subgroup of G and let

K = N(P)/C(P).

By the lemma, K is isomorphic to a subgroup of \text{Aut}(P). But since P is a cyclic group of order p_1, we have |\text{Aut}(P)|=p_1-1 and thus |K| \mid p_1 - 1. Hence |K| \mid \phi(n). Clearly |K| \mid n and so |K|=1 because \gcd(n, \phi(n))=1. So N(H) = C(H) and thus, by the Burnside’s normal complement theorem, there exists a normal subgroup Q of G such that

G=PQ, \ P \cap Q = \{1\}.

Thus |Q|=p_2 \cdots p_k < n and hence, by the induction hypothesis, Q is cyclic. Let

L = N(Q)/C(Q)= G/C(Q).

Again, by the lemma, L is isomorphic to a subgroup of \text{Aut}(Q). Also, since Q is cyclic,

|\text{Aut}(Q)|= (p_2-1) \cdots (p_k-1) \mid \phi(n).

Therefore |L| \mid \phi(n). But clearly |L| \mid n and thus |L|=1, i.e. G = C(Q). So Q is in the center of G and therefore G is abelian because P is abelian, Q is in the center of G and G=PQ. Hence G \cong P \times Q and so G is cyclic because both P and Q are cyclic with coprime orders. \Box

Remark 1. By the fundamental theorem of finite abelian groups, a group of square-free order is abelian if and only if it is cyclic. This result can also be used instead of the last line of the solution.

Remark 2. The converse of the above problem is also true, i.e. if n is a positive integer and the only group of order n is \mathbb{Z}_n, then \gcd(n, \phi(n))=1. The proof is by contradiction. First note that if there exists a prime divisor of n such that p^2 also divides n, then \mathbb{Z}_p \times \mathbb{Z}_{n/p} would be a group of order n which is not cyclic because \gcd(p,n/p)=p \ne 1. So we may assume that n is square-free. Now if \gcd(n,\phi(n)) > 1, then there exists prime divisors p,q of n such that p divides q-1. That implies the existence of a non-abelian group G of order pq and so G \times \mathbb{Z}_{n/pq} would be a non-abelian (hence non-cyclic) group of order n.