sep14 - = 3 * 49 has a normal Sylow 7-subgroup, so we shall...

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Math 5125 Wednesday, September 14 September 14, Ungraded Homework Construct a nonabelian group of order 75. The Sylow 5-subgroup of a group of order 75 is always normal, so we will let our subgroup H for the semidirect product be of order 25, specifically Z 5 × Z 5 , and we shall let K Aut ( H ) such that | K | = 3. To do this, we need to find an element k Aut ( H ) such that | k | = 3. One way to do this is to define k ( x , y ) = ( - y , x - y ) (where we have used additive notation in Z 5 × Z 5 ). Now set K = h k i and let φ : K Aut ( H ) denote the natural inclusion. Then H o φ K is a group of order | H || K | = 25 * 3 = 75, and is nonabelian because φ does not map K to 1. Specifically, (( 1 , 0 ) , k )(( 1 , 0 ) , 1 ) = (( 1 , 0 ) k ( 1 , 0 ) , k ) = (( 1 , 0 )+( 0 , 1 ) , k ) = (( 2 , 0 ) , k ) , yet (( 1 , 0 ) , 1 )(( 1 , 0 ) , k ) = (( 2 , 0 ) , k ) . Construct a nonabelian group of order 147 which has a cyclic Sylow 7-subgroup. By Sylow, a group of order 147
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Unformatted text preview: = 3 * 49 has a normal Sylow 7-subgroup, 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)....
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