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Math Help - characters of S_n from A_n

  1. #1
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    characters of S_n from A_n

    Using Frobenius Reciprocity, derive the character table of S_n from that of A_n when
    (a) n = 3
    (b) n = 4
    (c) n = 5
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  2. #2
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    Quote Originally Posted by akc2010 View Post
    Using Frobenius Reciprocity, derive the character table of S_n from that of A_n when
    (a) n = 3
    (b) n = 4
    (c) n = 5
    This is for n=5.

    The character table for S_5 is

    \begin{matrix}<br />
\text{size}&1 & 10 & 20 & 30 &24 & 15 &20\\<br />
&1 & (12) & (123) & (1234) & (12345) &(12)(34) & (12)(345)\\<br />
U & 1 & 1 & 1 & 1 & 1 &1 &1\\<br />
U' & 1 & -1 & 1 & -1 & 1 & 1 & -1\\<br />
V & 4 & 2 & 1 & 0 & -1 & 0 &-1\\<br />
V' & 4 & -2 & 1 & 0 & -1 & 0 & 1 \\<br />
\wedge^2V & 6 & 0 & 0 & 0 & 1 & -2 &0 \\<br />
W & 5 & 1 & -1 & -1 & 0 & 1&1\\<br />
W' & 5 & -1 & -1 & 1 &0 &1 & -1 \\<br />
\end{matrix}

    The character table for A_5 is here.
    I followed the notations of the book, Fulton's "Representation Theory : A first course".

    Note that the conjugacy class (12345) in S_5 with size 24 splits into two conjugacy classes (12345) and (21345) in A_5 with size 12, respectively. Note also that the conjugacy classes (12), (1234) and (12)(345) are not available in A_5.

    Let H=A_5 and G=S_5. Then, by using Frobenius's reciprocity

    Res U = U, Res U' = U, Ind U = U \oplus U',
    Res V = V, Res V' = V, Ind V = V \oplus V',
    Res W = W, Res W' = W, Ind W = W \oplus W', and
    Res \wedge^2V = Y \oplus Z, Ind Y = Ind Z = \wedge^2V.

    The last line is obtained by using the lemma,

    \chi^G(g) = \sum_{i=1}^s|G:H|\frac{k_i}{l}\chi(h_i),

    where l is the order of the conjugacy class in G of g,
    h_1, h_2, \cdots, h_s are representatives of s conjugacy classes of H,
    k_1, k_2, \cdots , k_s be the orders of these classes.
    Last edited by aliceinwonderland; December 15th 2009 at 10:21 PM.
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