Exam2-Solns - : G S 5 with ker H . | H | = 2 implies ker =...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
Exam 2 Solutions 1. (a) True. If | G | = p n then G has a normal subgroup of size p k for 1 k n . Thus G is simple IFF n = 1. ( { e } is defined to not be simple) (b) False. Counterexample: Let G = Z and K = 2 Z . Then G/K = Z 2 . Let ϕ : G G/K such that ϕ ( g ) = 0 g G . Kerϕ = G 6 = K . 2. (a) Not a group. There’s no identity. If a < 0 then a * b = | ab | 6 = a for any b G . (No identity also implies no inverses) (b) Not a group. The identity is 0, but does not have inverses. e.g. - 1 * b == 1 + b - b = - 1 6 = 0 for all b G . 3. H = { (1 , 1) , (0 , 2) , (1 , 3) , (0 , 0) } , H + (0 , 1) = { (1 , 2) , (0 , 3) , (1 , 0) , (0 , 1) } . 4. (a) e 2 = e e T T 6 = Ø. Let a,b T . Thus a m = e = b n for some m,n > 1. ( ab - 1 ) mn = ( a m ) n (( b - 1 ) n ) m (because G is abelian) = ( a m ) n (( b n ) - 1 ) m = e n e m = e . Thus ab - 1 T and therefore T is a subgroup by the 1-step subgroup test. (b) Suppose there exists aT G/T with o ( aT ) = m < . This implies a m T = eT = T a m T . Thus for some n > 1 ( a m ) n = e which implies a mn = e which in turn implies a T . Therefore aT = T and hence the only element of G/T of finite order is the identity. 5. By assumption there exists H G such that | H | = 2 and H is not normal. [ G : H ] = 5. By the generalized Cayley’s theorem there exists a homomorphism
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: : G S 5 with ker H . | H | = 2 implies ker = { e } or H . If ker = H then H C G , which would be a contradiction. Thus ker = { e } which implies is 1-to-1 and G = ( G ) S 5 . 6. | G | = 5 2 13 2 . n 5 1 (mod 5) and n 5 | 13 2 . Thus n 5 = 1 and the Sylow 5-subgroup, H , is normal. n 13 1 (mod 13) and n 13 | 5 2 . Thus n 13 = 1 and the Sylow 13-subgroup, K , is normal. H K = { e } since they are both p-groups for distinct p s. H,K C G and | HK | = 4225 = | G | thus HK = G and G = H K . Since H and K are of order p 2 they are both abelian. Thus G is also abelian. Hence G = Z 25 Z 169 , Z 5 Z 5 Z 169 , Z 25 Z 13 Z 13 , or Z 5 Z 5 Z 13 Z 13 by the Fundamental Theorem of Finite Abelian Groups....
View Full Document

This note was uploaded on 04/19/2011 for the course MATH 21373 taught by Professor Johnson during the Spring '11 term at Carnegie Mellon.

Ask a homework question - tutors are online