I am currently reading "A Course in Ring Theory" by Passman, and I know other questions about this proof have been asked here before, but I really want to understand the proof in the book.
The theorem states "Let $K$ be a field and let $G$ be a finite group. Then $K[G]$ is an Artinian ring that is Wedderburn if and only if the characteristic of $K$ dose not divide $|G|$."
I understand the direction wear he shows if the characteristic of $K$ divides the order of $G$ then $K[G]$ is not Wedderburn. For the other direction the author starts off with
"Conversely, assume that $|G| = n$ is not $0$ in $K$ and consider the $n \times n$ matrix ring $M_n(K)$, where we index the rows and columns by the elements of $G$. Then the map $\theta : K[G] \rightarrow M_n(K)$ given by $\theta(g) = (\delta_{y,xy})$ is easily seen to determine a $K$-algebra embedding, since $$\theta(g)\theta(h) = (\delta_{y,xg})(\delta_{y,xh}) = (\delta_{y,xgh}) = \theta(gh)$$ Here, of course, $g, h, x, y \in G$. Furthermore =, $\delta_{x, y} = 1$ when $x = y$ and $\delta_{x,y} = 0$ otherwise."
There is more to this direction of the proof, but My problem is I really don't understand the map $\theta$. I know $(\delta_{y,xg}) \in M_n(K)$, but know what it looks like.
I guess, it writes $\theta(g)=(\delta_{y,x{\pmb g}})$, and it's meant as the matrix with the $x,y$ (or $y,x$) entry $1$ if $y=xg$ and $0$ otherwise.
In other words, the matrix $\theta(g)$ is a permutation matrix which sends the basis element $x\in G$ of the $K$-vector space $K[G]$ to the other basis element $xg$, so this is just multiplying by $g$, which is a linear map in the group algebra.
Now, $\theta$ is defined on the basis $G$ of $K[G]$, and if we choose the correct directions (actually, you might rather need multiplication from left or by $g^{-1}$), then it maps multiplication on $G$ to matrix multiplication, as stated, and obviously it extends linearly to $K[G]$.