Question about a not semisimple group algebra Let $K$ be a field, $G$ a finite group and $K[G]$ a group algebra.
We have the following result:
Theorem: if the characteristic of $K$ divides the order of $G$, then $K[G]$ is not semisimple.
Let $I:=\{\sum_{g \in G} a_g.g: a_g \in K, \: \text{such that} \: \sum a_g=0 \}$ a two-sided ideal of $K[G]$.
The idea is to prove that for each ideal $I$ of $K[G]$, there is no left ideal $J$ of $K[G]$ such that $K[G]=I \oplus J$ (i.e. such that $I \cap J=\emptyset$ and $I+J=K[G]$).
Please can someone give an explanation of the following sentence.
If $K[G]=I \oplus J$ for some left-ideal $J$ of $K[G]$, then $\dim J=1$ so it is spanned by, say, $\alpha=\sum_g b_g.g$.
I have difficulties to understand the notion of dimension of the ideal $J$.
 A: Since $K[G]$ is a $K$-algebra, it follows that any ideal is in fact also a $K$-subspace (check this). So we can talk about the dimension of ideals in a $K$-algebra (as the dimension of the ideal as $K$-vector space).
Define
$$\omega: K[G] \to K: \sum_{g}k_g g \mapsto \sum_g k_g$$
Check that this is a $K$-algebra epimorphism. Note that $I= \ker \omega$ and from the isomorphism theorem we know that we have an isomorphism of $K$-algebra's
$$K \cong K[G]/I$$
Taking $K$-dimensions of both sides:
$$1=\dim_K (K) = \dim _K(K[G]) - \dim_K(I) = |G| - \dim_K(I)$$
and thus 
$$\dim_K(I) = |G| -1$$
Since  $K[G]= I \oplus J$, we also have $$|G| = \dim_K(K[G]) = \dim_K(I) + \dim_K(J) = |G| -1 + \dim_K(J)$$
and we deduced that $\dim_K(J) = 1$. Thus $J$ is $K$-spanned by one element.
A: The ideal $I$ you have defined is commonly called the augmentation ideal of the group ring, and it is the kernel of a ring homomorphism onto $K$.
Since the dimension of $K[G]$ is the finite cardinality of $G$, the rank theorem says that the dimension of $I$ as a $K$ vector space is $n-1$, and therefore its complement would have to be $1$ dimensional.

By the way, I find it much easier to show $K[G]$ is not semisimple by proving there is a nonzero central nilpotent element. This is necessarily in the Jacobson radical, and semisimple rings cannot have a nontrivial Jacobson radical.  One such central nilpotent element is $x=\sum_{g\in G}g$, which satisfies $x^2=|G|x$, and if the characteristic of $K$ divides $|G|$, the right side is zero.
