{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

all2 - Math 4124 Wednesday March 16 Fifth Homework...

Info icon This preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
Math 4124 Wednesday, March 16 Fifth Homework Solutions 1. Let G be a group of order 10 with a normal subgroup H of order 2. (a) Prove that H is contained in the center of G (hint: If 1 = x H and g G , then 1 = gxg - 1 H ). (b) Prove that G is abelian. (c) Let a G \ H . Prove that aH has order 5 in G / H . (d) Prove that G has an element y of order 5. (e) Let 1 = x H . Prove that the order of xy is neither 2 nor 5. (f) Prove that G is cyclic. (g) Prove that G = Z / 10 Z . (a) Since | H | = 2, it has two elements, one of which is the identity 1; we’ll call the other element x . Let g G . The g 1 g - 1 = 1. Also gxg - 1 H \ 1 (because H G ), hence gxg - 1 = x . It follows that g commutes with every element of H and we deduce that H is contained in the center of G . (b) Note that G / H is cyclic because it has order 5, which is a prime, and all groups of prime order are cyclic. Since H is contained in the center of G , we conclude that G is abelian. (c) By Lagrange’s theorem, | aH | divides | G / H | = 5. Since aH = 1 because a / H , we see that | aH | = 5. (d) Since | aH | = 5, we see that | a | = 1 or 2. Also | a | divides 10 by Lagrange’s theorem. Therefore | a | = 5 or 10. If | a | = 5, we’re finished. On the other hand if | a | = 10, then | a 2 | = 5 as required. (e) Note that | x | = 2. Since G is abelian, ( xy ) 2 = x 2 y 2 = y 2 = 1 and ( xy ) 5 = x 5 y 5 = x = 1. Therefore | xy | = 2 or 5. (f) Note that xy = 1. Therefore | xy | = 1, 2 or 5. It follows from Lagrange’s theorem that | xy | = 10 and we conclude that G is cyclic. (g) All cyclic groups of order 10 are isomorphic to Z / 10 Z . The result follows, be- cause G is cyclic of order 10. 2. 3.3.8 on page 101. Let p be a prime and let G be the group of p -power roots of 1 in C . Prove that the map z z p is a surjective homomorphism. Deduce that G is isomorphic to a proper quotient of itself. Define θ : G G by θ g = g p . Then for x , y G , we have ( xy ) p = x p y p because G is abelian, consequently θ ( xy ) = ( xy ) p = x p y p = θ x θ y ,
Image of page 1

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
so θ is a homomorphism. Suppose g G , then g = e 2 k π i / p n for some k , n Z . Set h = e 2 k π i / p n + 1 . Then h G and θ h = g . Therefore θ is surjective. Finally e 2 π i / p ker θ , so ker θ = 1. The result now follows from the first isomorphism theorem. 3. Let G be a group, and suppose there is a homomorphism of G onto S 3 (the symmetric group of degree 3) with kernel K . Determine the number of subgroups of G which contain K , and show that exactly three of these subgroups are normal. By the subgroup correspondence theorem, the subgroups containing K are in a one- to-one correspondence with the subgroups of G / K , and this correspondence preserves normality. This means we have to determine the subgroups of S 3 . Any subgroup of S 3 has order dividing 6, so any subgroup not equal to 1 or S 3 has prime order and is therefore cyclic. It follows easily that the subgroups of S 3 are 1, S 3 , ( 12 ) , ( 23 ) , ( 31 ) , ( 123 ) , so there are six subgroups of G which contain K . Finally the normal subgroups of S 3 are 1, S 3 and ( 123 ) . We conclude that the number of normal subgroups containing K is 3, as required.
Image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern