According to [Wikipedia](http://en.wikipedia.org/wiki/Regular_representation): 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)



Number of irreducible representations
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](http://books.google.co.uk/books?id=oWzoOWp_tWgC&pg=PA76&lpg=PA76&dq=regular+representation+of+a+group+ring&source=bl&ots=qo3ZDIsInc&sig=TYRFfwKXCKQdjX83IGH9aeVlzvk&hl=en&ei=Bk33SqmPLYKj4Qbc7ozeAw&sa=X&oi=book_result&ct=result&resnum=10&ved=0CDEQ6AEwCQ#v=onepage&q=regular%20representation%20of%20a%20group%20ring&f=false). A simple dimension counting argument should convince you that you have at least one ideal for each conjugacy class.


