I am currently reading a research paper and have encountered a point that I am struggling to understand. In the paper, it is proven that $J(F_5[D_{30}])^5 = (0)$, indicating that the Jacobson radical of the group algebra $F_5[D_{30}]$ raised to the power of $5$ yields the zero element. Consequently, the subgroup $1+J(F_5[D_{30}])$ possesses an exponent of $5$ and a nilpotency class that is less than or equal to $4$. Additionally, it is established that $a^3-1 = 4+a^3 \in J(F_5[D_{30}])$, leading to the conclusion that $4b+a^9b$, $4a^2+a^8$, and $4ab+a^{10}b$ also belong to $J(F_5[D_{30}])$. In this context, $D_{30}=\{{a,b\mid a^{15}=b^2=1, bab=a^{-1}}\}$ denotes the dihedral group of order $30.$ Here, $\mathbb{F}_{5}$ is a field of characteristic $5$, consisting of $5$ elements.
However, I am uncertain about the derivation of the expressions $4b+a^9b$, $4a^2+a^8$, and $4ab+a^{10}b$. I understand that $J(F_5[D_{30}])$ serves as an ideal of the group algebra $F_5[D_{30}]$, allowing us to assert that $4b+a^3b$, $4a^2+a^5$, and $4ab+a^4b$ belong to $J(F_5[D_{30}])$ by multiplying the element $4+a^3$ on the right side with $b$, $a^2$, and $ab$, respectively. However, I am unsure about the origin of the terms $a^9$, $a^8$, and $a^{10}$.
If you could provide further clarification on this matter, it would be greatly appreciated.
I assume that as we can start with generator $a$ and prove that $4+a^3\in J$ we can do the same with the generator $a^2$ and prove that $4+a^6\in J$.
Premultiply by $b$ and get $b4+ba^6=4b+a^9 b\in J$.
Premultiply by $a^2$ and get $4a^2+a^8\in J$.
Premultiply by $ab$ and get $ab4+aba^6=4ab+a^{10} b\in J$.