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# Adv Alegbra HW Solutions 75 - 1 4 2 3 1 2 3 4 1 3 24 1 4 2...

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75 Solution. Write C = C G ( x ) . Since H has index 2, we have H C G , by Proposition 2.97, and so the second isomorphism the- orem gives | CH || C H |=| C || H | ; by Lagrange s theorem, [ H : C H ]=[ CH : C ] . The left side is | x H | ,for C H = C H ( x ) , by part (i). But 2 = [ G : H ]=[ G : CH ][ CH : H ] implies [ CH : H ]= 1o r [ CH : H ]= 2, and this gives the result. 2.123 Prove that the group UT ( 3 , I 2 ) in Example 2.150 is isomorphic to D 8 . Solution. You may use the fact that the only nonabelian groups of order 8 are D 8 and Q . 2.124 (i) How many permutations in S 5 commute with ( 12 )( 34 ) , and how many even permutations in S 5 commute with ( 12 )( 34 ) . Solution. If β commutes with α = ( 12 )( 34 ) , then β( 5 ) = 5. Hence, the problem has been reduced to looking within S 4 . Since we cannot use centralizers and conjugates, there is no choice but to look at the list of 24 elements of S 4 , checking which commute with ( 12 )( 34 ) . The answer is: ( 1 ), ( 12 )( 34 ), ( 13 )( 24
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Unformatted text preview: 1 4 )( 2 3 ), ( 1 2 ), ( 3 4 ), ( 1 3 24 ), ( 1 4 2 3 ). (Note that if γ = ( 1 3 2 4 ) , then γ 2 = α .) Only the f rst four of these are even. (ii) How many permutations in S 7 commute with ( 1 2 )( 3 4 5 ) ? Solution. There are 1 2 7 · 6 × 1 3 5 · 4 · 3 = 230 conjugates of α in S 7 , because two permutations in a symmetric group are conjugate if and only if they have the same cycle structure. Now S 7 acts on itself by conjugation; the orbit of α is its conjugacy class, and its stabilizer is its centralizer C S 7 (α) . By the orbit-stabilizer theorem, [ S 7 : C S 7 (α) ] = 230. Therefore, | C S 7 (α) | = 7 ! / 230 = 12. (iii) Exhibit all the permutations in S 7 commuting with ( 1 2 )( 3 4 5 ) ....
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