Updated below, but still not sure of my approach. I am trying to solve problem $13.6$ from Isaacs' Algebra, a Graduate Course. I am probably not understanding the question, but here is my attempt. A proper solution would be most appreciated. I copied the problem to an image for reference.
$\textbf{13.6}$ Let $G$ be a finite $p$-group and $F$ a field of characteristic $p$. Construct the group algebra $A=FG$. Show that $J(A)$ has codimension one in $A$. (In other words, $\dim(J(A))=|G|-1$.)
HINT: Work by induction on $|G|$. If $|G|>1$, let $Z\subseteq\mathbf{Z}(G)$ with $|Z|=p$, and let $1\ne z\in Z$. Show that $A/(1-z)A\cong F(G/Z)$ and that $(1-z)A\subseteq J(A)$.
NOTE: The augmentation ideal of $A$ (see the note following Problem $12.8$) has codimension one and so is a maximal right ideal and thus contains (and therefore equals) $J(A)$. Therefore $1-g\in J(A)$ for all $g\in G$. Observe that
Using the hint I use induction starting with the fact that dimension of $ J(F(G/Z)) = |G|/p-1$
I want to show that the dimension of $J(A)=J(F(G))$ is $|G|-1$
The surjective homomorphism $\theta:F(G)\rightarrow F(G/Z)$ given by $$\sum a_g\cdot g \rightarrow \sum a_g \cdot Zg $$
then provides (Corrollary 13.2) that $J(F(G/Z)) = \theta (J(F(G)))$
Since the kernel of $\theta$ is $(1-z)A$ I know that $F(G/Z) \cong A/(1-z)A$ and that $\theta (J(A)) =J(F(G/Z)))$ so $J(F(G/Z)))=J(A)/(1-z)A$
Assuming this much is correct (I am very new at this) it seems that I need to determine the dimension of the ideal $(1-z)A$. Then I get the dimension of $J(A)$ from the equation (which I need confirmation of) $\dim(X/Y)=\dim(X)-\dim(Y)$.
I am not seeing easily what this dimension is. I think a spanning set is the set $\{1-z)g\}$ but I'm not sure how to determine the number of independent elements present. I do know (or believe) that the set of elements of $G$ is a spanning set, with the proviso that the sum of the coefficients is zero. So an unrestricted spanning set (basis) would be the set $\{1-g\}$. But this doesn't seem to get me where I am supposed to be. I also think that the elements of $(1-z)A$ are nilpotent, and thus $(1-z)A \subset J(A)$. This does seem to imply that the dimension of $J(A)$ equals $|G|-1$. But I don't think this requires the induction proof.
UPDATE: I see that the kernel is bigger than $(1-z)A$, containing $(1-y)A$ for all $1 \neq y \in Z$. However, I am still not set up to calculate the requested dimensions. My approach is obviously clumsy.
Thank you in advance for clarifying. I am trying to learn from the book.