This preview shows page 1. Sign up to view the full content.
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 1to1 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 5subgroup, H , is normal. n 13 1 (mod 13) and n 13  5 2 . Thus n 13 = 1 and the Sylow 13subgroup, K , is normal. H K = { e } since they are both pgroups 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.
 Spring '11
 Johnson
 Algebra

Click to edit the document details