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Under these assumptions, note that $\mathrm{Rad}(kG)$ coincides with the augmentation ideal $\omega(kG)$ of $kG$. Now, as remarked by Jeremy Rickard, the question is equivalent to whether the elements $(g-1)^s$ with $g\in G$ generate $\mathrm{Rad}^s(kG)$ as an ideal of $kG$. Theorem 3.7 in Section 5.3 of the book [D.S. Passman: The algebraic structure of group rings] provides an explicite description of $Rad(kG)^s$ for all $s>0$. As proved in the beautiful David's answer, $Rad(kG)^s$ is in fact generated by the elements of the form $(g-1)^p$ with $g\in G$ in the given situation.

Under these assumptions, note that $\mathrm{Rad}(kG)$ coincides with the augmentation ideal $\omega(kG)$ of $kG$. Now, as remarked by Jeremy Rickard, the question is equivalent to whether the elements $(g-1)^s$ with $g\in G$ generate $\mathrm{Rad}^s(kG)$ as an ideal of $kG$. Theorem 3.7 in Section 5.3 of the book [D.S. Passman: The algebraic structure of group rings] provides an explicite description of $Rad(kG)^s$ for all $s>0$. As proved in the beautiful David's answer, $Rad(kG)^s$ is in fact generated by the elements of the form $(g-1)^s$ with $g\in G$ in the given situation.

Under these assumptions, note that $\mathrm{Rad}(kG)$ coincides with the augmentation ideal $\omega(kG)$ of $kG$. Now, as remarked by Jeremy Rickard, the question is equivalent to whether the elements $(g-1)^s$ with $g\in G$ generate $\mathrm{Rad}^s(kG)$ as an ideal of $kG$. However, according to Theorem 3.7 in Section 5.3 of the book [D.S. Passman: The algebraic structure of group rings], this does not necessarily seem to be the case.

Under these assumptions, note that $\mathrm{Rad}(kG)$ coincides with the augmentation ideal $\omega(kG)$ of $kG$. Now, as remarked by Jeremy Rickard, the question is equivalent to whether the elements $(g-1)^s$ with $g\in G$ generate $\mathrm{Rad}^s(kG)$ as an ideal of $kG$. Theorem 3.7 in Section 5.3 of the book [D.S. Passman: The algebraic structure of group rings] provides an explicite description of $Rad(kG)^s$ for all $s>0$. As proved in the beautiful David's answer, $Rad(kG)^s$ is in fact generated by the elements of the form $(g-1)^p$ with $g\in G$ in the given situation.

Note that in this case $\mathrm{Rad}(kG)$ coincides with the augmentation ideal $\omega(kG)$ of $kG$. Now, as remarked by Jeremy Rickard, the question is equivalent to whether the elements $(g-1)^s$ with $g\in G$ generate $\mathrm{Rad}^s(kG)$ as an ideal of $kG$. However, according to Theorem 3.7 in Section 5.3 of the book [D.S. Passman: The algebraic structure of group rings], this does not necessarily seem to be the case.

Under these assumptions, note that $\mathrm{Rad}(kG)$ coincides with the augmentation ideal $\omega(kG)$ of $kG$. Now, as remarked by Jeremy Rickard, the question is equivalent to whether the elements $(g-1)^s$ with $g\in G$ generate $\mathrm{Rad}^s(kG)$ as an ideal of $kG$. However, according to Theorem 3.7 in Section 5.3 of the book [D.S. Passman: The algebraic structure of group rings], this does not necessarily seem to be the case.