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 orbitstabilizer 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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 Fall '11
 KeithCornell
 Symmetric group, α, S7 commute, centralizer C S7

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