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Unformatted text preview: MAT 322: Algebra with Galois Theory Professor Claus Sorensen Problem Set 4 Rik Sengupta October 16, 2009 1. As on page 61 in GT, just above Theorem 4.32, make a table of all possi ble cycle structures for S 5 , and find the cardinalities of all the conjugacy classes in S 5 , together with their parities. Solution. We have: Partition Element Number in Conjugacy Class Parity 1+1+1+1+1 1 1 even 1+1+1+2 (ab) 10 odd 1+1+3 (abc) 20 even 1+2+2 (ab)(cd) 15 even 1+4 (abcd) 30 odd 2+3 (ab)(cde) 20 odd 5 (abcde) 24 even There are a number of ways of checking this, but the most obvious ways are to check that adding the numbers in the conjugacy classes gives us  S 5  = 120, and to check  A 5  = 60 as well. Both these conditions hold true in the table. 2. Let G be a group with  G  = 12 , and assume it contains more than two elements of order three. Show that G is isomorphic to A 4 . Solution. We have,  G  = 12 = 2 2 3. Now, we know  Syl 3 ( G )  divides 4, and  Syl 3 ( G )  1(mod 3) from the Third Isomorphism Theorem, so that  Syl 3 ( G )  = 1 or  Syl 3 ( G )  = 4. But since the assumption is that there are more than two elements of order three, we must have  Syl 3 ( G )  = 4. Define a homomorphism as the conjugate action : G Sym(Syl 3 ( G )) S 4 . Then,  G : N G ( P )  =  Syl 3 ( G )  = 4, and  G : P  = 4 for P Syl 3 ( G ), so P = N G ( P ). 1 Hence, ker( ) = \ P Syl 3 ( G ) N G ( P ) = \ P Syl 3 ( G ) P = { e } . Now,  ( G )  =  G  = 12 and  S 4  = 24, so  S 4 : ( G )  = 2. If ( acb ) ( G ), then ( acb ) 2 = ( abc ) ( G ), so ( G ) has every 3cycle, and the 3cycles generate A 4 . Hence, ( G ) = A 4 . Thus, G = G/ ker( ) A 4 , by the First Isomorphism Theorem. 3. Let G be a group with  G  = 30 . Show that G is not simple. Solution. We have,  G  = 30 = 2 3 5. So, from Sylows Third Theorem and the corollary to Sylows Second Theorem, we know  Syl 5 ( G )  divides 6 and  Syl 5 ( G )  1(mod 5), so that either  Syl 5 ( G )  = 1 or  Syl 5 ( G )  = 6. Similarly,  Syl 3 ( G )  divides 10 and  Syl 3 ( G )  1(mod 3), so that either  Syl 3 ( G )  = 1 or  Syl 5 ( G )  = 10. If  Syl 3 ( G )  = 10 and  Syl 5 ( G )  = 6, then there are 20 nontrivial elements of order 3 and 24 nontrivial elements of order 5, which is a contradiction because  G  = 30. So,  Syl p ( G )  = 1 for at least one value of p { 3 , 5 } . But if  Syl p ( G )  = 1, then by the corollary to Sylows Second Theorem, P C G for some P Syl p ( G )  , and hence G is definitionally not simple. So, if  G  = 30, then G is not simple. 4. Inside GL n ( C ) , consider the subgroup B consisting of all upper triangular invertible matrices. Compute the derived series of B ....
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This note was uploaded on 05/07/2010 for the course MAT 322 at Princeton.
 '09
 CLAUSSORENSEN
 Algebra

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