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
$\displaystyle \begin{matrix}
\text{size}&1 & 10 & 20 & 30 &24 & 15 &20\\
&1 & (12) & (123) & (1234) & (12345) &(12)(34) & (12)(345)\\
U & 1 & 1 & 1 & 1 & 1 &1 &1\\
U' & 1 & -1 & 1 & -1 & 1 & 1 & -1\\
V & 4 & 2 & 1 & 0 & -1 & 0 &-1\\
V' & 4 & -2 & 1 & 0 & -1 & 0 & 1 \\
\wedge^2V & 6 & 0 & 0 & 0 & 1 & -2 &0 \\
W & 5 & 1 & -1 & -1 & 0 & 1&1\\
W' & 5 & -1 & -1 & 1 &0 &1 & -1 \\
\end{matrix}$
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 $\displaystyle \oplus$ U',
Res V = V, Res V' = V, Ind V = V $\displaystyle \oplus$ V',
Res W = W, Res W' = W, Ind W = W $\displaystyle \oplus$ W', and
Res $\displaystyle \wedge^2V$ = Y $\displaystyle \oplus$ Z, Ind Y = Ind Z = $\displaystyle \wedge^2V$.
The last line is obtained by using the lemma,
$\displaystyle \chi^G(g) = \sum_{i=1}^s|G:H|\frac{k_i}{l}\chi(h_i)$,
where $\displaystyle l$ is the order of the conjugacy class in G of g,
$\displaystyle h_1, h_2, \cdots, h_s$ are representatives of s conjugacy classes of H,
$\displaystyle k_1, k_2, \cdots , k_s$ be the orders of these classes.