Unformatted text preview: = 3 * 49 has a normal Sylow 7subgroup, so we shall let H = Z 49 . We need to ﬁnd an automorphism k of H which has order 3. One possibility is the map deﬁned 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 φ : K → Aut ( H ) be the natural inclusion, then H o φ K is a group of order  H  K  = 49 * 3 = 147, and it is nonabelian because φ is nontrivial (same argument as in the previous problem)....
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This note was uploaded on 01/02/2012 for the course MATH 5125 taught by Professor Palinnell during the Fall '07 term at Virginia Tech.
 Fall '07
 PALinnell
 Math, Algebra

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