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Unformatted text preview: Math 5125 Monday, August 22 August 22, Ungraded Homework Exercise 3.1.36 on page 89 Prove that if G / Z ( G ) is cyclic, then G is abelian. Write Z = Z ( G ) . If G / Z is cyclic, then G / Z = xZ for some x ∈ G . This means that the left cosets of Z in G are of the form x a Z for some a ∈ Z . Therefore every element of G is of the form x a z for some z ∈ Z . Suppose we have another such element, say x b w . Then ( x a z )( x b w ) = zwx a x b = wzx b x a = ( wx b )( zx a ) , consequently any two elements of G commute and we have proven that G is abelian. Exercise 3.2.4 on page 95 Show that if G = pq for some primes p and q (not necessarily distinct), then either G is abelian or Z ( G ) = 1. By Lagrange’s theorem,  Z ( G )  = 1 , p , q or pq . If  Z ( G )  = pq , then G is abelian, so we may assume without loss of generality that  Z ( G )  = p . But then  G / Z ( G )  = q . Since groups of prime order are cyclic, we deduce from the previous exercise that G is abelian and the result follows. Let G be a finite group, let H ≤ G and let K G . Prove that  HK : H  =  K : K ∩ H  . Note that HK and K ∩ H are subgroups of G (the former by, for example, the second iso morphism theorem), so the result to be proven makes sense. By the second isomorphism theorem, HK / K ∼ = H / H ∩ K . Since everything in sight is finite, we see that  HK  /  K  =  H  /  H ∩ K  and hence  HK  /  H  =  K  /  K ∩ H  . The result follows. Math 5125 Wednesday, August 24 August 24, Ungraded Homework Exercise 3.3.3 on page 101 Prove that if H is a normal subgroup of G of prime index p , then for all K ≤ G either (i) K ≤ H or (ii) G = HK and  K : K ∩ H  = p . We know that HK ≤ G , because H , K ≤ G and one of them is normal. Obviously HK ⊇ H . Also  G / H  = p and p is prime, so by Lagrange’s theorem G / H has only two subgroups, namely H / H and G / H . By subgroup correspondence theorem, we now see that HK = H or G . If K is not contained in H , then we cannot have HK = H and we deduce that HK = G . By the second isomorphism theorem we have HK / H ∼ = K / K ∩ H , consequently  G / H  =  K / K ∩ H  and the result follows. Let H G be groups such that G / H ∼ = Z / 3 Z × Z / 3 Z . Prove that G has at least four normal subgroups of index 3. Let H G such that G / H ∼ = Z 3 × Z 3 . We need to prove that G has at least 4 normal sub groups of index 3. It is easily checked that Z 3 × Z 3 has 4 subgroups of order 3. Since Z 3 × Z 3 has order 9, these subgroups will all have index 9/3 = 3. Also all these subgroups are normal because Z 3 × Z 3 is an abelian group. Thus G / H has 4 normal subgroups of index 3. It now follows from the subgroup correspondence theorem that G has 4 normal subgroups of index 3 containing H , which proves the result....
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 Fall '07
 PALinnell
 Algebra, Sets, Normal subgroup, Abelian group, Subgroup, Sylow psubgroup

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