Find a corresponding subrepresentation of a irreducible representation in a representation. Say we know an irreducible representation $\tau$ appears $k$ times in a  representation $\rho$. How do we find the corresponding subrepresentations (i.e. the k many G-stable subspaces) in $\rho$?
Example: Consider the regular representation of $S_3$ and we know the standard representation appears twice, so how do we find the subspaces? 
 A: Let $G$ be a finite group, $V$ be an irreducible representation of $G$ over $\mathbb{C}$, and $W$ be any other representation. Then define the $V$-isotypic component $W[V]$ to be the sum of all subrepresentations of $W$ which are isomorphic to $V$:
$$ W[V] = \sum_{X \subseteq W, X \cong V} X.$$
In order to compute $W[V]$, you can use the character $\chi_V: G \to \mathbb{C}$ of $V$. Specifically, let $\psi_V \in \mathbb{C}[G]$ be the group algebra element
$$ \psi_V = \frac{\dim V}{|G|} \sum_{g \in G} \chi_V(g^{-1}) g.$$
I claim that $\psi_V$ acts as a projection $W \to W[V]$ for any representation $W$. Firstly, it certainly belongs to the centre of $\mathbb{C}[G]$, since
$$ \sum_{g \in G} \chi_V(g^{-1}) hgh^{-1} = \sum_{g \in G} \chi_V((h^{-1} g h)^{-1}) g = \sum_{g \in G} \chi_V(g^{-1}) g.$$
Next, let $U$ be any irreducible representation of $G$. Since $U$ is irreducible, the element $\psi_V$ acts on $U$ as a scalar, so we can learn this scalar by computing the trace 
$$ \operatorname{tr}(\psi_V|_U) = \frac{\dim V}{|G|} \sum_{g \in G} \chi_V(g^{-1}) \chi_U(g) = (\dim V)\langle \chi_V, \chi_U \rangle$$
which is zero if $V \not\cong U$, and $\dim V$ if $V \cong U$. This proves the result:

The isotypic component $W[V]$ is equal to the image of $\psi_V$ acting on $W$.

Let's do the example with $G = S_3$, $W = \mathbb{C}[S_3]$ the regular representation, and $V$ the two-dimensional irreducible representation. The character of $V$ is $\chi_V(e) = 2$, $\chi_V((123)) = \chi_V((213)) = -1$, and $\chi_V(g) = 0$ for the transpositions. Hence
$$ \psi_V = \frac{1}{3} \left( 2e - (123) - (213) \right).$$
So the image of $\psi_V$ inside $W = \mathbb{C}[G]$ is spanned by
$$ \begin{aligned}
3 \psi_V e &= 2e - (123) - (213) \\
3 \psi_V (123) &= 2(123) - (213) - e \\
3 \psi_V (213) &= 2(213) - e - (123) \\
3 \psi_V (12) &= 2(12) - (13) - (23) \\
3 \psi_V (13) &= 2(13) - (23) - (12) \\
3 \psi_V (23) &= 2(23) - (12) - (13)
\end{aligned}$$
and it is easy to see that the first three vectors span a two-dimensional subspace, and so do the second three vectors.
