According to Wikipedia: If G is a finite group and K is the complex number field, the regular representation is a direct sum of irreducible representations, in number at least the number of conjugacy classes of G.
Can anyone prove this? (Only an honors level student, so please try to keep it as simple as possible)
If I remember correctly, that statement can be proven via the equivalence between the (group) representations of $G$ and the (algebra) representations of the group algebra $K[G]$. If $K=\mathbb{C}$ is the field of complex numbers (or any other field of characteristic 0, or in general if $\textrm{char} K$ does not divide the order of $G$) then by Maschke's Theorem $K[G]$ is semisimple, and moreover it is Artinian (by finite-dimensionality) so by Artin-Wedderburn it decomposes (as a module over itself) as a direct sum of matrix rings over some division ring over $K$. If $K=\mathbb{C}$, there are no finite dimensional division rings over $\mathbb{C}$ so each summand is a ring of matrices over the complex numbers. The irreducible representations of the group are in one to one correspondence with left ideals of this ring, which are well known (look for instance here. A simple dimension counting argument should convince you that you have at least one ideal for each conjugacy class.
It's not entirely trivial, but here's the sketch of the proof if you would like to finish it yourself. Note that people usually denote complex numbers by $\mathbb C$:
(1) Every irreducible representation comes as part of regular representation, denoted $\mathbb C[G]$
To prove this, take any non-trivial representation $R$ and consider $\mathrm{Hom}\,(R, \mathbb C[G])$. A careful examination should show it's non-trivial as well.
(2) The character $\chi_R$ is a function on $G$ defined as $\chi _ R(g) = \mathrm{tr}\,g| _ R$. Prove that value of any character on an element of $G$ depends only on element's conjugacy class.
This immediately follows from $\mathrm{tr}\,A = \mathrm{tr}\,BAB^{-1}$ (property of trace).
(3) Prove that any function on $G$ that depends only on a conjugacy class is a linear combination of some characters $\chi_R$.
I'm not sure I remember how to prove this in an elementary way, but it's doable.
(4) Prove your statement from (1)-(3).
Indeed, there must be at least as many irreducible representations as there are conjugacy classes and all of them appear in regular representation by (1), so you have your statement.
There are many references on representation theory, you can basically pick up any book to answer your question. I would like to point out the following very accessible text if you are be interested in learning more:
