I am going to present a statement and a proof. The proof makes use of idempotents which makes it a little cumbersome.
Is there a proof that does not use idempotents?
(using well-known theorems is OK even if their proofs do use idempotents)
Statement:
Let $k=\mathbb{F}_p$. Let $G$ be a finite group. Consider the group ring $kG$. Let $I$ be the augmentation ideal of $kG$. Denote the Jacobson radical of $kG$ by $J=J(kG)$. Let $U$ be the indecomposable projective $kG$-module such that $U/UJ$ is isomorphic to the trivial $kG$-module $k$. Write $kG=U\oplus W$ ($W$ is the direct sum of the other indecomposable modules in the decomposition of $kG$).
Then $I=UJ\oplus W$.
My proof:
The augmentation map $\epsilon:kG\rightarrow k$ factors through $kG/J$ because $J\subset I$ because $kG/I$ is simple. Write $\epsilon=\epsilon'\circ\pi$ (that is, $\epsilon$ factors as $kG\overset{\pi}{\longrightarrow}kG/J\overset{\epsilon'}{\longrightarrow}k$). Let $U'=\ker\epsilon'$ and let $W'\subset kG/J$ be a direct complement of $U$' in $kG/J$. That is $kG/J=U'\oplus W'$. Write $U'=e_1kG/J$ and $W'=e_2kG/J$ with $e_1,e_2$ orthogonal idempotents. Since $J$ is a nil ideal (even nilpotent), we can lift $e_1,e_2$ to a pair of orthogonal idempotents $f_1,f_2\in kG$. Write $U=f_1kG$ and $W=f_2kG$. Then $kG=U\oplus W$.
Now, it's easy but a little cumbersome to prove that $W \subset I$ and $U \cap I = UJ$. Thus $I=UJ\oplus W$.
Finally, we have to our $U$ is the same $U$ as in the statement. It is enough to show that $U/UJ=k$. This is true because $UJ=U \cap J$ and thus $U/UJ=U/(U\cap J)=(U+J)/J$. That last expression equals $k$, but to show it I need to use idempotents and again it's easy but a little cumbersome.
Roughly, $kG$ only has one part that can surject onto $k$; the rest is in the kernel.
In more detail: $kG$ is a direct sum of projective indecomposable modules $P_i$ each with a simple top factor $P_i/JP_i = S_i$ (with $d_i = \left|\{ j : S_i \cong S_j\}\right|$ the dimension of $S_i$). Since $k$ has dimension 1, there is only one such p.i.m. with top quotient $k$, and it is usually called $P_1$.
Consider the restriction of $\epsilon$ to each $P_i$. The kernel of the restriction certainly contains $JP_i$, but for $i > 1$ it must also contain the top factor too, since $S_i \not\cong S_1 = k$.
The kernel contains $JP_1 \oplus P_2 \oplus \ldots \oplus P_n$ but the quotient by that subgroup is already $k$, so the kernel of $\epsilon$ must be exactly $JP_1 \oplus P_2 \oplus \ldots \oplus P_n$. Here $U=P_1$ and $W=P_2 \oplus \ldots \oplus P_n$.