What are the possible dimensions of an irreducible representation of a group of order 21?
I'm not sure what theorems might be useful for solving this problem. I know that every element of the group has order either 1, 3, or 7, and that there is at least one element of order 3 and one of order 7 (because not every element can have order p where $p\in \{21,3,7\}$, there is an identity element, and there is an element of order either 3 or 7). The dimension of the representation is by definition $\chi(1)$ where $\chi$ is the character of the representation. I know that the group is completely reducible as it is finite, though I'm not sure if this helps.
I shall talk in terms of character degrees. We denote by $\text{Irr}(G)$ the set of irreducible characters of $G$. I will prove that, indeed, the irreducible characters of $G$ are all linear or $G$ has $3$ linear characters and $2$ irreducible characters of degree $3$.
If $G$ is abelian then necessarily $\chi(1)=1$ for each $\chi\in\text{Irr}(G)$.
If $G$ is not abelian, since $\chi(1)$ divides $|G|$ for every $\chi\in\text{Irr}(G)$ and $\sum_{\chi\in\text{Irr}(G)}\chi(1)^2=|G|=21$ then necessarily $\chi(1)\in\{1,3\}$. In this case, there exists at least one $\chi\in\text{Irr}(G)$ with $\chi(1)=3$, as $G$ is nonabelian. The complex conjugate of an irreducible character is an irreducible character. Hence, $\overline\chi\in\text{Irr}(G)$. To prove the statement it suffices to check that $\chi\not=\overline\chi$.
Suppose that $\ker\chi\not=1$. Then it holds that $G/\ker\chi$ is a group of prime order. Now, since $\chi\in\text{Irr}(G/\ker\chi)$ and $G/\ker\chi$ is abelian it follows that $\chi(1)=1$, which is a contradiction. Thus, $\ker\chi=1$.
If $\chi=\overline\chi$ then for every $g\in G$ it would hold $$\chi(g)=\overline\chi(g)=\chi(g^{-1}).$$ Since $\ker\chi=1$, this yields that $g=g^{-1}$ and hence $g^2=1$ for all $g\in G$. But this implies that $G$ is abelian, which is a contradiction. Hence $\chi\not=\overline\chi$ and we are done.