Question about isotypical components Consider $V=\bigotimes^3(\mathbb{C}^2)$ as a $\mathfrak{S}_3$ representation. One of its isotypical component is $S^3(\mathbb{C}^2)$, which is a linear subspace of symmetric tensors of $\bigotimes^3(\mathbb{C}^2)$. Since $\mbox{dim} V=8$, there should be atleast one other isotypical components. I need to describe all possible isotypical components. 
The hint that is given is: $\mathfrak{S}_3$ contains the cyclic group $C_3$ as a normal subgroup of index two. 
I don't understand how this can be useful. 
Any hint is appreciated.  
 A: The symmetric group $\mathfrak{S}_3$ has three irreducible representations (in characteristic not dividing $6$). These are the trivial representation, the sign representation, and its $2$-dimensional reflection representation. As you observed, $S^3(\mathbb{C}^2)$ is the four-dimensional isotypic component of $T^3(\mathbb{C}^2)=(\mathbb{C}^2)^{\otimes 3}$ for the trivial representation. The sign representation does not occur for dimension reasons (there are no anti-symmetric $n$-tensors on a vector space of dimension less than $n$), so the remaining four dimensions must be the isotypic component for the reflection representation. 
Using the hint, you can obtain this decomposition explicitly as follows: for a tensor $a \otimes b \otimes c$ I will write $[a \otimes b \otimes c]^j$ for its symmetrization into the eigenspace of eigenvalue $e^{2 \pi i j/3}$ for the $3$-cycle $(123)$ generating $C$. Explicity, $(123)(a \otimes b \otimes c)=c \otimes a \otimes b$ and so
$$[a \otimes b \otimes c]^j=a \otimes b \otimes c+e^{-2 \pi i j/3} c \otimes a \otimes b+e^{2 \pi i j /3} b \otimes c \otimes a. $$ 
Now since $(12) (123) (12)^{-1}=(123)^{-1}$, we know a priori that the action of $(12)$ on $T^3(\mathbb{C}^2)$ will fix the eigenspace $E^0$ for eigenvalue $1$ and permute the eigenspaces $E^1$ and $E^2$ for the eigenvalues $e^{2 \pi i j/3}$ and $e^{-2 \pi i j/3}$. Writing $e_1,e_2$ for the standard basis of $\mathbb{C}^2$, these are
$$E^0=\mathbb{C} \{[e_1 \otimes e_1 \otimes e_1]^0, [e_2 \otimes e_2 \otimes e_2]^0, [e_1 \otimes e_1 \otimes e_2]^0, [e_1 \otimes e_2 \otimes e_2]^0 \} $$
$$E^1=\mathbb{C} \{  [e_1 \otimes e_1 \otimes e_2]^1, [e_2 \otimes e_2 \otimes e_1]^1 \} $$ and 
$$E^2=\mathbb{C} \{  [e_1 \otimes e_1 \otimes e_2]^2, [e_2 \otimes e_2 \otimes e_1]^2 \}. $$
With our notation we have for instance $(12)([e_1 \otimes e_1 \otimes e_2]^1)=[e_1 \otimes e_1 \otimes e_2]^2$ so that the span of $[e_1 \otimes e_1 \otimes e_2]^1$ and $[e_1 \otimes e_1 \otimes e_2]^2$ is one of the two copies of the two dimensional reflection representation, the other being spanned by $[e_2 \otimes e_2 \otimes e_1]^1$ and $[e_2 \otimes e_2 \otimes e_1]^2$. Evidently $E^0$ is precisely the space of symmetric tensors.
