Has every infinite simple group a faithful irreducible representation?

Question

Has every infinite simple group a faithful irreducible representation?

This question solves the finite case. However, the proof requires a non-trivial linear representation of a finite group. I want to know if the conclusion is true for an infinite simple group. If it is not, can you give me a counterexample?

Answer 1

The statement is true for infinite groups with some exceptions if you allow infinite-dimensional representations

Unless $G$ is a p-group, where $p$ is the characteristic of $k$, the group ring $k[G]$ is not local, so there exist at least two isomorphism classes of simple modules, in particular, there is an irreducible representation where $G$ acts nontrivially.

Such a representation is faithful, because $G$ is simple and so any nontrivial homomorphism from $G$ is injective.

Conversely, if $G$ is $p$-group (such as a Tarski monster), where $p$ is the characteristic of $k$, then $k[G]$ is local and hence there is only one irreducible representation, the trivial one, which is not faithful.

Answer 2

In GROUPS WITH NO NONTRIVIAL LINEAR REPRESENTATIONS by A.J. BERRICK. Example 1.3 gives a simple, locally finite group which has only trivial linear representation. Thus it has no faithful irreducible linear representation.

I searched on Google for some relevant results and found it eventually by hours, may it be helpful to the late-comers.

Answer 3

When a ring has a faithful simple right module it is called right primitive. A faithful simple $R[G]$ module yields a faithful irreducible representation, so looking at when $R[G]$ is right primitive might be fruitful.

At one time, right primitive group rings (other than obvious ones like the semisimple ones) were scarce. You might be interested in seeing the following paper that observed how to get a class of more interesting examples:

Formanek, Edward, and Robert L. Snider. "Primitive group rings." Proceedings of the American Mathematical Society 36.2 (1972): 357-360.

A better known theorem is that $R[G]$ is prime if $R$ is prime and $G$ has no nontrivial finite normal subgroups. Right primitive rings are prime, so it is a necessary condition for $R[G]$ to be right primitive. Requiring $G$ to be simple is sufficient to say $R[G]$ is prime, but it's not enough to get primitivity.