Given $G$ is a group of order $12$ admitting an irreducible $3-$dimensional representaion. What are the dimensions of its irreducible representaions?
Is there a theorem that gives an answer? I am guessing it is using the group order, $12$.
Is it true that their dimensions divide the group order? So they could be $12, 6, 4, 3, 2, 1$, this would fit into the fact that G has one irreducible representation of dimension $3$
And can we know how many there are, given the information from the question?....
If $V_1,V_2\cdots V_n$ are all the irreducible representations (up to isomorphism) of $G$, where $n$ is the number of conjugacy classes of $G$, then $$\sum_{i=1}^n\dim(V_i)^2 =|G|.$$
We are given that $G$ has a representation of dimension $3$, whose square $9$ is three short of $12$, the order of the group $G$. There is only one way to write $3$ as a sum of squares, that is $1+1+1$ so by the above formula it must be that there are three representations of dimension $1$ in addition to the given representation.