Image of augmentation ideal of a group ring under ring homomorphism. Let $G,H$ be two finite $p$-groups and $f:\mathbb{F}_p[G]\to \mathbb{F}_p[H]$ a ring homomorphism of mod $p$ group rings.
Is it always true that $f(I_G)\subset I_H$, where $I_G,I_H$ are the respective augmentation ideals?
Since the group rings are local this is equivalent to $f(I_G)\neq \mathbb{F}_p[H]$ or $1 \notin f(I_G)$.
 A: First argumentation.
Let $$ be any field and let $A$ be a $$-algebra.

*

*Every homomorphism of $$-algebras $ε$ from $A$ to $$ is automatically surjective, whence its kernel is a two-sided ideal of codimension $1$.
It follows that $A =  1 ⊕ \ker(ε)$, whence $ε$ is uniquely determined by its kernel.


*For every two-sided ideal $I$ of $A$ of codimension $1$ the composite $ \to A \to A / I$ is an isomorphism of $$-algebras (since it is an isomorphism on the level of $$-vector spaces).


*Consequently, we have a one-to-one correspondence between homomorphism of $$-algebras from $A$ to $$ and two-sided ideals of $A$ of codimension $1$.
If $A$ is local, then it follows that there exists at most one homomorphism of $$-algebras from $A$ to $$.
Therefore, if $A$ and $B$ are two augmented $$-algebras with $A$ local, then every homomorphism of $$-algebras $f$ from $A$ to $B$ is already a homomorphism of augmented $$-algebras, i.e., it satisfies the following equivalent conditions:
$$
  ε_B ∘ f = ε_A \,,
  \quad
  f^{-1}(I_B) = I_A \,,
  \quad
  f(I_A) ⊆ I_B \,.
$$
Second argumentation.
In the given situation, the augmentation ideal $I_G$ is nilpotent.
This entails that every element of $I_G$ is nilpotent.
It follows that $f(I_G)$ consists of nilpotent elements, whence $1 ∉ f(I_G)$.
