I am preparing for my algebra qualifying exam, and this is one of the past exam problems, which I got stuck on. I would appreciate any help!
Let $F=\mathbb{Z}/11\mathbb{Z}$, the field of $11$ elements. Let $G=\langle x:x^{11}=1\rangle$ be the group of $11$ elements, with the group operation multiplication. Then we can consider the group algebra $F[G]$, and for an element $r\in F[G]$, we can consider the $F$ linear map $T_r:F[G]\to F[G]$ given by $a\to ra$ for $a\in F[G]$.
The problem asks to find the Jordan canonical form of $T_r$.
What I got so far is that $F[G]\cong \mathbb{Z}/11\mathbb{Z}[x]/(x^{11}-1)=\mathbb{Z}/11\mathbb{Z}[x]/(x-1)^{11}$, and that $F[G]$ is a finitely generated $F[t]$ module, hence by the structure theorem, we may let $F[G]\cong F[t]/(q_1(t))\oplus\cdots\oplus F[t]/(q_d(t))$ where $q_i(t)\in F[t]$ are prime powers.
From here, I am not sure where I could go. If $r=1\in F[G]$ so that $T_r$ is the identity map, then $t-1\in F[t]$ is the minimal polynomial of $T_r$ and hence $F[G]\cong [F[t]/(t-1)]^{11}\cong F^{11}$. On the other hand, this seems weird to me because we would then have $\mathbb{Z}/11\mathbb{Z}[x]/(x-1)^{11}\cong (\mathbb{Z}/11\mathbb{Z})^{11}$?