# Perfect Groups with faithful complex irreps

Is it true that a perfect finite group $$G$$ has a faithful complex irrep if and only if the center of $$G$$ is cyclic?

The corresponding fact is true for quasisimple finite $$G$$

Does every quasisimple finite group have a faithful complex irrep?

And certainly it is necessary to have cyclic center for a finite group to have a faithful complex irrep.

So the question is really: Does every perfect group with cyclic center have a faithful complex irrep?

An example of a non perfect group with cyclic center which has no faithful complex irrep is given here

https://mathoverflow.net/q/57129/387190

No. Take PerfectGroup(245760,4) (this requires GAP 4.12 :-) ), generated by

(1,11,10,8,14)(2,4,12,15,16)(3,13,5,6,7)(17,25,28,21,23)(18,19,24,30,29)(20,27,22,26,31)(33,45,39,41,35)(34,47,42,36,46)(38,40,48,44,43),
(1,13,12,16,9)(2,7,5,6,14)(3,10,15,8,4)(17,21,23,27,30)(18,24,32,22,28)(19,20,29,25,31)(33,40,45,36,37)(34,38,48,39,41)(35,46,42,44,47),
(1,14,2)(3,12,16)(4,9,7)(5,15,8)(6,13,10)(17,31,18)(19,28,29)(20,25,23)(21,32,24)(22,30,26)(33,39,38)(34,43,47)(35,36,45)(37,42,48)(41,46,44)

It has a structure $$(2^4\times 2^4\times 2^4):A_5$$, where $$2^4$$ is the irreducible $$A_5$$-module of dimension 4 that does not come from the permutation representation. It has 21 minimal normal subgroups (all of type $$2^4$$). The solvable radical $$(2^4)^3$$ is composed from 21 classes of order 15 (which each generate one of the minimal normal subgroups) and 63 classes of order 60 (that each generate a normal subgroup of order 256), and of course the identity.

This group has trivial center (thus cyclic) but has no faithful irreducible representation.

This is the smallest possible example, and the only one of order 245760. (There are further ones in order 491520, not neccessarily just extensions with this one.)

The criterion of Gaschütz for (a finite group) $$G$$ having a faithful irreducible complex representation described in the linked mathoverflow post can be used to show theoretically that examples like the one of ahulpke must exist.

The criterion for the existence is that the socle of $$G$$ should be generated by a single conjugacy class of $$G$$.

Consider groups $$G$$ with a normal elementary abelian subgroup $$N$$ of order $$p^n$$ such that $$S = G/N$$ is nonabelian simple. Then conjugacy classes in $$G$$ have size at most $$|S|$$, so if $$n > |S|$$ then there can be no such representation.

For all such $$S$$ and all primes $$p$$ we can find examples of the type, by letting $$M$$ be any nontrivial irreducible module for $$S$$ over $${\mathbb F}_p$$, and setting $$G = M^k \rtimes S$$ for any $$k>0$$, with the induced module action of $$S$$ on $$M^k$$.

The example of ahulpke is of this type with $$n=12$$ and $$M$$ of dimension $$4$$.

• Ok this seems very interesting I like that it's more conceptual. Let me see if I'm following. I'm identifying $M$ as $2^4$ the 4 dimensional irreducible submodule of the standard permutation representation of $A_5$ on $2^5$. Then $M^k \rtimes A_5$ is a perfect group with trivial center for any $k$(your answer math.stackexchange.com/questions/4438581/…) why for $k=3$ are there are no faithful irreps but for $k=2$ there are? You claim for sufficiently large $k$ (say $k=15$ i.e. $n=15(4)=60$) there are no irreps. Is $k=3$ just an accident? Nov 23, 2022 at 15:02
• Careful: $A_5$ has two irreducible representations in dimension 4 over $\mathbb{F}_2$, and in my example it is the one that does not come from the permutation module. (This is not to say that it must fail, just that it will be a different group.) Nov 23, 2022 at 16:56
• @IanGershonTeixeira As ahulpke has pointed out, you have the wrong $M$. Sorry I have no answer to your "why?" questions! Nov 23, 2022 at 17:07
• And, just to make it more strange, the example I gave turns out to be the only one of order 245760. So it will work for the one 4-dimensional module, but not the other one. Nov 23, 2022 at 17:14