# Possible degrees of group characters

Let $$\varphi : G \to \text{GL}(m, \mathbb{C})$$ be an irreducible representation. Its degree, or equivalently the degree of its (linear) character, is by definition $$m$$.

I was wondering whether the following is true: if $$m$$ is any (non-zero) integer, does there exist a finite group $$G$$ and an irreducible representations $$\varphi: G \to \text{GL}(m, \mathbb{C})$$? In other words, can every integer occur as the degree of an irreducible representation?

If the answer is yes, is there also a set of groups and representations $$\{G_i, \varphi_i\}_{i\in\mathbb{N}}$$ known for which $$\varphi_i$$ has degree $$i$$?

References to source material are very welcome.

Yes, for instance the symmetric group $$S_{n+1}$$ has an irreducible representation of degree $$n$$. Precisely, take the canonical permutation representation $$V$$ of $$S_{n+1}$$; it has degree $$n+1$$, and we can write $$V=V'\oplus L$$ where $$L$$ is a copy of the trivial representation and $$V'$$ is irreducible, and of degree $$n$$. It is called the standard representation of $$S_{n+1}$$.
Let $$G_p$$ represent an extraspecial $$p$$-group of order $$p^3,$$ $$p$$ prime. There are two non-isomorphic types of such groups, but that does not matter: the character degrees (over $$\mathbb{C}$$) are the same: $$G_p$$ has $$p^2$$ linear characters and $$p-1$$ irreducible characters of degree $$p$$. That last fact is of use, for if $$m=p_1^{r_1} \cdot p_2^{r_2} \cdots p_k^{r_k}$$ is the decomposition in primes, and for a group $$G$$, $$G^{(i)}$$ denotes the direct product of $$i$$ copies of $$G$$, and $$A$$ is an arbitrary finite abelian group, then the group $$G_{p_1}^{(r_1)} \times G_{p_2}^{(r_2)} \cdots \times G_{p_k}^{(r_k)} \times A$$ has an irreducible character of degree $$m$$. In fact, it has $$\Pi_{i=1}^k (p_i-1)^{r_i}$$ irreducible characters of degree $$m$$. The order of this group is $$m^3 \cdot |A|$$.
Note If $$m$$, and for linear characters $$1$$, are the only occurring irreducible characters degrees of $$G$$, then it can be shown that at least one of the following occurs:
(a) $$G$$ has an abelian normal subgroup of index $$m$$
(b) $$m=p^e$$, $$p$$ prime and $$G$$ is the direct product of a $$p$$-group and an abelian group.