Nature of the subgroup generated by an unfaithful irreducible representation of a finite group

Question

An unfaithful irreducible representation of a finite group forms a subgroup of the group. For example, for the alternating group $A_4$, there is an unfaithful singlet representation which forms the cyclic group, $Z_3$ ($Z_3$ is a subgroup of $A_4$). Another example will be the symmetric group $S_4$ and its unfaithful doublet representation which forms the dihedral group $D_6$ ($D_6$ is a subgroup of $S_4$). Is such a subgroup somehow different from the other subgroups of the group? Can the group be written as some kind of a product involving such a subgroup?

Answer 1

As pointed out in the comment an unfaithful representation of a finite group forms a quotient of the group, not necessarily a subgroup.

The question is then are these quotients somehow different? The answer is yes.

An excellent description of this is given in this MO post.

My preffered terminology for the subgroup of $G$ generated by minimal normal subgroups of $G$ is the socle of $G$, but it is called the base of $G$ in the post.

Rephrasing the finite case in my preferred terminology, if $G$ is finite then it has an irreducible faithful representation over $\mathbb{C}$ if and only if the socle of $G$ is generated by a single class of conjugates in $G$.

As we are talking about quotients, I'm afraid the question about products doesn't make much sense.