Intersection of kernels of all irreducible representations is trivial

Statement
Suppose $$G$$ is a finite group and $$K$$ is a field whose characteristic does not divide the order of $$G$$. Consider all the irreducible linear representations of $$G$$ over $$K$$. Each such irreducible representation is a homomorphism to a general linear group and we can consider its kernel, which is a normal subgroup of $$G$$.

The intersection of all these subgroups is the trivial subgroup of $$G$$.

Note that it is not necessary that $$K$$ be a splitting field for $$G$$. However, we do require that the characteristic of $$K$$ not divide the order of $$G$$, although there are some examples where the result holds even when the characteristic of $$K$$ divides the order of $$G$$.

Related facts

 * Intersection of kernels of all indecomposable linear representations is trivial: This holds in general even when the characteristic of the field divides the order of the group.

Proof outline
The key idea is:


 * 1) Consider the regular representation, a permutation representation corresponding to the left-regular group action of the group on itself. This is a faithful linear representation.
 * 2) Decompose this as a sum of irreducible linear representations. This decomposition is possible because of Maschke's averaging lemma, and the actual decomposition can be worked out by the orthogonal projection formula, though that is not necessary for our purpose. (it turns out that every irreducible linear representation appears as a summand of the regular representation; however, that is not necessary for the present proof).
 * 3) The intersection of the kernels of the irreducible representations being summed is the kernel of the regular representation, which is trivial. Hence, there is a collection of irreducible representations whose kernels intersect trivially. Thus, the intersection of kernels of all irreducible representations is trivial.