Dimension of irreps of $C_3v$ The $C_{3v}$ point group (symmetry group of a regular triangle) has 6 elements: $A$,$B$,$C$,$E$,$D$,$F$ (3 reflections, identity, 2 rotations). It has 3 conjugate classes $\phi_1=\{E\}$, $\phi_2=\{A,B,C\}$ and $\phi_3=\{D,F\}$. A reducible representation $R(g)$ with $2\times 2$ real matrices can be given and the character of the conjugate classes are 
$$\chi^R(\phi_1)=2$$ 
$$\chi^R(\phi_2)=0$$
$$\chi^R(\phi_3)=-1.$$
It follows from the orthogonality theorem that the square of the dimensions of irreducible representations equals the order of the group. In this case $1^2+1^2+2^2=6$. How do I see that two irreps have dimension 1 while the other has dimension 2?
 A: This is the dihedral group of order $6$, and it is well-known that it is isomorphic to the symmetric group $S_3$. The two representations of dimension $1$ are the trivial representation and the sign representation, whereas the representation of dimension $2$ is the standard representation, given by restricting the natural permutation action on the coordinates of $\mathbb{R}^3$ to the invariant subspace $x_1+x_2+x_3=0$.
A: We can do this without really thinking about what the group is.
If $G$ is a non-abelian group of order $6$ then $G$ has $3$ irreducible representations and their degrees are $1$, $1$ and $2$.
To see this, we first note that not all the irreducible representations can have degree $1$ since then the group would be abelian. So at least one has degree at least $2$. But as you noted yourself, the sum of the squares of the degrees equals the order of the group, so no irrep can have degree $3$ or more. And we cannot have two of degree $2$ as this would give order at least $8$, which leads to the conclusion that the only possible degrees are precisely $1$, $1$ and $2$.
