# Let $G$ be a non-abelian finite group whose all non-linear irreducible characters are faithful. Is there a classification of these groups?

Let $$G$$ be a non-abelian finite group whose all non-linear irreducible characters are faithful. Is there some kind of a classification of these groups?

If $$G$$ is a finite non-abelian simple group, the above is certainly true. The symmetric groups $$S_n$$ also has the above property. I can see that for a group $$G$$ as above, if $$N\neq \{1\}$$ is a normal subgroup, then $$[G,G]\leq N$$. In other words, $$G/N$$ is abelian. The converse is also true, that is, if $$G/N$$ is abelian for all normal subgroups $$\{1\} \neq N\leq G$$, then all non-linear irreducible characters are faithful.

So in other words, I am looking for a classification of non-abelian finite groups whose non-trivial quotients are all abelian.

Thanks in advance for any kind of reference or help.

• Comments have been moved to chat; please do not continue the discussion here. Before posting a comment below this one, please review the purposes of comments. Comments that do not request clarification or suggest improvements usually belong as an answer, on Mathematics Meta, or in Mathematics Chat. Comments continuing discussion may be removed. Commented May 24 at 0:36

A buzzword here is "just nonabelian". In general group is called just $$X$$ if it has property $$X$$ but all its proper quotients do not have property $$X$$. Therefore a group is just nonabelian if it is nonabelian but all its proper quotients are abelian.
Let $$G$$ be a finite just nonabelian group. As discussed in the comments section, $$G'$$ is characteristically simple. If $$G'$$ is nonabelian then $$G' \cong S^n$$ for some nonabelian simple group $$S$$. The centralizer $$C_G(G')$$ does not contain $$G'$$, so it must be trivial, and therefore $$G$$ embeds in $$\mathrm{Aut}(S^n) \cong \mathrm{Aut}(S) \wr S_n$$. Thus $$G$$ is an extension by $$G' \cong S^n$$ of an abelian subgroup $$A$$ of $$\mathrm{Out}(S^n) \cong \mathrm{Out}(S) \wr S_n$$. Moreover $$A$$ acts transitively on the $$n$$ factors of $$S$$ in $$G'$$. I am not sure if one can say more -- perhaps.
If $$G'$$ is abelian then $$G$$ is metabelian, so $$G$$ is just (strictly) metabelian. These groups were classified by Newman in the references below. It makes a big difference whether $$Z(G)$$ is trivial or not. If $$Z(G)$$ is trivial, the first paper shows that $$G \le \mathrm{AGL}_1(q)$$ for some prime power $$q$$. If $$Z(G)$$ is nontrivial then $$G' \le Z(G)$$, so $$G$$ is a class-2 nilpotent group, and in fact a $$p$$-group for some $$p$$ with $$Z(G)$$ cyclic and $$G' \cong C_p$$. These groups are classified in the second paper. Examples include extraspecial $$p$$-groups and the modular $$p$$-group $$C_{p^{n-1}} \rtimes C_p$$.
• This a really good answer. Just one note for people like me who were confused for a second: $C_G(G')$ is the centralizer of a normal subgroup hence normal itself and since all nontrivial normal subgroups of $G$ have to contain $G'$ for the condition to hold this means $C_G(G')$ is trivial when $G'$ is nonabelian. Commented May 17 at 6:34