HW 04 - MAT 322: Algebra with Galois Theory Professor Claus...

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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 3-cycle, and the 3-cycles 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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HW 04 - MAT 322: Algebra with Galois Theory Professor Claus...

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