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By A. I. Kostrikin, I. R. Shafarevich

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Let a E N. Then 4sH(a)(bH) = abH = ba'H = bH since b-gab = a' E N C H. Therefore, N C Ker(4iH) and Ker(4H) is the largest normal subgroup of G contained in H. 3, we will indicate how to use this result to prove the existence of normal subgroups of certain groups. 4) Corollary. Let H be a subgroup of the finite group G and assume that IGI does not divide [G : HJ!. Then there is a subgroup N C H such that N 0 {e} and N a G. Proof. Let N be the kernel of the permutation representation 4iH. 3 N is the largest normal subgroup of G contained in H.

Let p and q be primes such that p > q and let G be a group of order pq. (1) If q does not divide p - 1, then G Z. (2) If q I p - 1, then G Zpq or G Zp x m Z. when 0: Zq -' Aut(Zp) = Z. is a nontrivial homomorphism. All nontrivial homomorphisms produce isomorphic groups. Proof. 7) G has a subgroup N of order p and a subgroup H of order q, both of which are necessarily cyclic. 6). Since it is clear that N n H = (e) and NH = G, it follows that G is the semidirect product of N and H. The map 0 : H - Aut(N) given by Oh (n) = hnh is a group homomorphism, so if q does not divide I Aut(N)I = I Aut(Zp)l = 1Z;1 = p - 1, then 0 is the trivial homomorphism.

10) Examples. (1) 2Z = {even integers} is a subring of the ring Z of integers. 2Z does not have an identity and thus it fails to be an integral domain, even though it has no zero divisors. (2) Z,,, the integers under addition and multiplication modulo n, is a ring with identity. Z has zero divisors if and only if n is composite. Indeed, ifn=rs for 1

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