Section 1.7. Group Actions 71 that is, if p is not a divisor of n, then np−1 1 mod p. Multiplying both sides by n, we have np n mod p, a congruence also holding when p is a divisor of n this is Fermat’s Theorem. We have seen, in Proposition 1.70, that A4 is a group of order 12 having no subgroup of order 6. Thus, the assertion that if d is a divisor of |G|, then G must have a subgroup of order d, is false. However, this assertion is true when G is a p-group. Proposition 1.116. If G is a group of order p , then G has a normal subgroup of order pk for every k . Proof. We prove the result by induction on 0. The base step is obviously true, and so we proceed to the inductive step. By Theorem 1.113, the center of G is a nontrivial normal subgroup: Z(G) = {1}. Let Z Z(G) be a subgroup of order p as any subgroup of Z(G), the subgroup Z is a normal subgroup of G. If k , then pk−1 p −1 = |G/Z|. By induction, G/Z has a normal subgroup H∗ of order pk−1. The Correspondence Theorem says there is a subgroup H of G containing Z with H∗ = H/Z moreover, H∗ G/Z implies H G. But |H/Z| = pk−1 implies |H| = pk, as desired. Abelian groups (and the quaternions) have the property that every subgroup is normal. At the opposite pole are groups having no normal subgroups other than the two obvious ones: {1} and G. Definition. A group G is called simple if G = {1} and G has no normal subgroups other than {1} and G itself. Proposition 1.117. An abelian group G is simple if and only if it is finite and of prime order. Proof. If G is finite of prime order p, then G has no subgroups H other than {1} and G otherwise Lagrange’s Theorem would show that |H| is a divisor of p. Therefore, G is simple. Conversely, assume that G is simple. Since G is abelian, every subgroup is normal, and so G has no subgroups other than {1} and G. Hence, if x G and x = 1 (simple groups are nontrivial), then x = G. If x has infinite order, then all the powers of x are distinct, and so x2 x is a forbidden subgroup of x , a contradiction. Therefore, every x G has finite order. If x has (finite) order m and m is composite, say m = k, then xk is a proper nontrivial subgroup of x , a contradiction. Therefore, G = x has prime order. Corollary 1.118. A finite p-group G is simple if and only if |G| = p. Proof. We claim that if |G| p, then G is not simple. If G is abelian, then Proposition 1.117 shows that G is not simple. Hence, we may assume that G is not abelian that is, its center, Z(G), is a proper subgroup. By Theorem 1.113, we have Z(G) = {1}. But Z(G) G, and so G is not simple.
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