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

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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 α : 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

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Hence, ker( α ) = \ P Syl 3 ( G ) N G ( P ) = \ P Syl 3 ( G ) P = { e } . Now, | α ( G ) | = | G | = 12 and | S 4 | = 24, so | S 4 : α ( G ) | = 2. If ( acb ) α ( G ), then ( acb ) 2 = ( abc ) α ( G ), so α ( G ) has every 3-cycle, and the 3-cycles generate A 4 . Hence, α ( G ) = A 4 . Thus, G = G/ ker( α ) 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 Sylow’s Third Theorem and the corollary to Sylow’s 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 Sylow’s 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 . Solution. Consider B GL n ( C ) and choose X, Y B . Let A = XY X - 1 Y - 1 . Then, we have, det( A ) = det( XY X - 1 Y - 1 ) = det( X )det( Y )det( X - 1 )( Y - 1 ) = 1 . But det( A ) = n Y i =1 a ii , so A only has 1s on its main diagonal. Hence, the first derived subgroup B (1) is generated by matrices of the form 1 * . . . * 0 1 * . . . . . . . . . . . . * 0 0 . . . 1 . 2
Now, choose P, Q B (1) . A rather long computation shows that R = PQP - 1 Q - 1 has 1s on the main diagonal and 0s on the diagonal above the main diagonal. Hence, B (2) consists of upper-triangular matrices of the form 1 0 * . . . * 0 1 0 * .

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

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