For a finite group and a field of characteristic p, the group algebra is local exactly when the group is a p-group

Suppose first that $G=P$ is a finite $p$-group. The quotient $k[P]\mathrm{/}{I}_P$ is the field $k$, so $I_P$ is a maximal ideal by [F1]. By [L1], $I_P$ is nilpotent. Every maximal left ideal of $k[P]$ therefore contains $I_P$, so the Jacobson radical from The Jacobson radical of a finite-dimensional algebra is the intersection of its maximal left ideals is exactly $I_P$. Hence $k[P]\mathrm{/}J(k[P])\cong k$ is simple, so $k[P]$ is local.

Conversely, assume $k[G]$ is local and suppose that $G$ is not a $p$-group. Choose a prime $q\mathrm{\ne }p$ dividing $\mathrm{\mid }G\mathrm{\mid }$. By [L3], $G$ contains a subgroup $C$ of order $q$. Since $q$ is invertible in $k$, the element $e_C:=q^{-1}{\sum }_{c\in C}[c]\in k[G]$ satisfies $e_C^2=e_C$ by [L4]. It is neither $0$ nor $1$: its coefficient at every $c\in C$ is the nonzero scalar $q^{-1}$, and either $C\mathrm{\ne }G$ or the coefficients at the nonidentity elements still differ from those of $1=[1_G]$.

A finite-dimensional local algebra has no nontrivial idempotent. Indeed, if $e^2=e$ with $e\mathrm{\ne }0,1$, then the nonzero proper left ideals $Ae$ and $A(1-e)$ are contained in maximal left ideals. Locality puts both inside the unique maximal left ideal, but $e+(1-e)=1$ then lies in that proper ideal, a contradiction. Applying this to $e_C$ contradicts step 1.2. Therefore every prime divisor of $\mathrm{\mid }G\mathrm{\mid }$ is $p$, and $G$ is a $p$-group.

Steps 1.1, 1.2, and 2.1 prove the equivalence. The trivial group is included in the forward direction with $I_1=0$.