70 Chapter 1. Groups I of the terms in the class equation other than |Z(G)|, and so p | |Z(G)| as well. Therefore, Z(G) = {1}. • McLain gave an example of an infinite p-group G with Z(G) = {1} (Robinson, A Course in the Theory of Groups, p. 362). Corollary 1.114. If p is prime, then every group G of order p2 is abelian. Proof. If G is not abelian, then its center Z(G) is a proper subgroup, so that |Z(G)| = 1 or p, by Lagrange’s Theorem. But Theorem 1.113 says that Z(G) = {1}, and so |Z(G)| = p. The center is always a normal subgroup, so that the quotient G/Z(G) is defined it has order p, and hence G/Z(G) is cyclic. This contradicts Exercise 1.77 on page 59. • Example 1.115. Who would have guessed that Cauchy’s Theorem (if G is a group whose order is a multiple of a prime p, then G has an element of order p) and Fermat’s Theorem (if p is prime, then ap ≡ a mod p) can be proved simultaneously? The elementary yet ingenious proof of Cauchy’s Theorem is due to McKay in 1959 (Montgomery–Ralston, Selected Papers in Algebra, p. 41) Mann showed me that McKay’s argument also proves Fermat’s Theorem. If G is a finite group and p is prime, denote the cartesian product of p copies of G by Gp, and define X = {(a0,a1,...,ap−1) ∈ Gp : a0a1 · · · ap−1 = 1}. Note that |X| = |G|p−1, for having chosen the last p − 1 entries arbitrarily, the 0th entry must equal (a1a2 · · · ap−1)−1. Introduce an action of Ip on X by defining, for 0 ≤ i ≤ p − 1, [i](a0,a1,...,ap−1) = (ai,ai+1,...,ap−1,a0,a1,...,ai−1). The product of the entries in the new p-tuple is a conjugate of a0a1 · · · ap−1: aiai+1 · · · ap−1a0a1 · · · ai−1 = (a0a1 · · · ai−1)−1(a0a1 · · · ap−1)(a0a1 · · · ai−1). This conjugate is 1 (for g−11g = 1), and so [i](a0,a1,...,ap−1) ∈ X. By Corol- lary 1.108, the size of every orbit of X is a divisor of |Ip| = p since p is prime, these sizes are either 1 or p. Now orbits with just one element consist of a p-tuple all of whose entries ai are equal, for all cyclic permutations of the p-tuple are the same. In other words, such an orbit corresponds to an element a ∈ G with ap = 1. Clearly, (1, 1,..., 1) is such an orbit if it were the only such, then we would have |G|p−1 = |X| = 1 + kp for some k ≥ 0 that is, |G|p−1 ≡ 1 mod p. If p is a divisor of |G|, then we have a contradiction, for |G|p−1 ≡ 0 mod p. We have thus proved Cauchy’s Theorem: if a prime p is a divisor of |G|, then G has an element of order p. Choose a group G of order n, say, G = In, where n is not a multiple of p. By Lagrange’s Theorem, G has no elements of order p, so that if ap = 1, then a = 1. Therefore, the only orbit in Gp of size 1 is (1, 1,..., 1), and so np−1 = |G|p−1 = |X| = 1 + kp

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