This preview shows page 1. Sign up to view the full content.
Unformatted text preview: = 3 * 49 has a normal Sylow 7subgroup, so we shall let H = Z 49 . We need to nd an automorphism k of H which has order 3. One possibility is the map dened by k ( h ) = h 30 . Since ( 30 , 49 ) = 1, we see ker k = 1, so k Aut ( H ) . Also one can check that k 3 ( h ) = h (an inelegant way to show this is to note that 30 3 1 mod 49, in fact 30 3 = 551 * 49 + 1). Thus h k i Aut ( H ) and h k i = 3. If we let : K Aut ( H ) be the natural inclusion, then H o K is a group of order  H  K  = 49 * 3 = 147, and it is nonabelian because is nontrivial (same argument as in the previous problem)....
View Full
Document
 Fall '07
 PALinnell
 Math, Algebra

Click to edit the document details