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**Example text**

B) Suppose that for any elements h E H and k E K holds hk = kh. Since K is abelian and H is abelian, G = H K is abelian. This contradicts the fact that the commutator subgroup K of G is of order q. This proves (b). ( c ) First we claim that H is not normal in G. Otherwise, for any h E H and k E K , hkh-lk-l E H n K = { e } and so hk = kh. By Sylow Theorem, - 36 the number of pSylow subgroups of G is greater than 1 and divides q . So it is Again by Sylow Theorem, plq - 1. q, 1200 (a) Let p , q be primes, p > q > 2 .

For any g E G, we have 9Pg-l C gKg-' 2 K since K is normal in G. Hence gPg-l is also a Sylow subgroup of K . By Sylow Theorem, there exists h E K such that gPg-l = h P h - l . Hence h-'gP(h-'g)-' = P and h-'g E N . So we have g = h . h-'g E K N . Thus we have G = K . N . - 1215 Let P be a Sylow subgroup of a finite group G. Let N = { x E G I xPx-' = P } . Let H be asubgroup of G, H 2 N . Prove: If y E G such that y H y - l = H , then y E N. (Indiana) 42 Solution. Obviously, P is a Sylow subgroup of H .

We can find integers u and v such that Iu - pl 5 1/2 and lw - vI 5 1/2. Then + + + a = b(p+vG) = b [ ( u + p- u ) + (v + v - v)Q] = bq+r where Q = u +v Q is in Z[-] and r = a - bq = b(p - u ) + b(v - v)Q. Obviously T E Z[-] and 6(T) = 6 ( b ) * ( l p - 2112 5 6(b). + 21v - (a + t) 2* wl2) < 6(b). Hence Z [ m is Euclidean. b) By a), Z [ G ] is a unique factorization domain. (1, -1) is the set of units of Z[-]. Suppose (20,yo) be an integer solution to the equation y2 2 = x3. In the ring Z [ G ] , we have (yo Q)(yo - G) = zg.