Regular representation I am stuck at the following question and dont know where to begin:
Let $\rho $ be the permutation representation associated to the operation of $D_3$ of order 6 on itself by conjugation. Decompose the character of $\rho$ into irreducible characters.
I was thinking since its a permutation matrix, we can think of it as 6 x 6 identity matrix and assign each row to the element of $D_3$ and take two elements and conjugate them and we get the permutation representation of specific element. But I am not sure how to proceed
 A: Let $D_3$ be the dihedral group of order 6. Write $D_3 = \langle a,b~|~a^3 = b^2 = 1, bab^{-1} = a^2\rangle$
The character table for $D_3$ is as follows (listed by conjugacy class):
\begin{array}{|c|c|c|c|}
 \hline & 1 & a & b \\ \hline \chi_1&1 &1 &1 \\ \hline \chi_2&1 &1 &-1\\ \hline \chi_3&2 &-1 &0\\ \hline \end{array}
So these are our three irreducible characters. 
The permutation representation is given by six $3 \times 3$ matrices. 
$\rho(1) = \left( \begin{array}{ccc}
1& 0 & 0 \\
0 & 1 &0  \\
0 & 0& 1 \end{array} \right) $  $\rho(a) = \left( \begin{array}{ccc}
0& 1 & 0 \\
0 & 0 &1  \\
1 & 0& 0 \end{array} \right) $  $\rho(b) = \left( \begin{array}{ccc}
0& 1 & 0 \\
1 & 0 &0  \\
0 & 0& 1 \end{array} \right) $  
The other three matrices are obtained by the appropriate multiplication.
Observing the trace of each of these three representative matrices, we get the permutation character: $\begin{array}{|c|c|c|c|}
 \hline & 1 & a & b \\\hline \chi_{per}& 3 & 0 & 1 \\\hline\end{array}$
We can then see that $\chi_{per} = \chi_1 + \chi_3$.
