Question

For any subgroup H of a group G, show that kH ∩ U(kG) ⊆ U(kH) and kH ∩ rad kG ⊆ rad kH.

Answer 1

Let $R = kH \subseteq S = kG$, and fix a coset decomposition $G = \bigcup_{i \in I} H \sigma_i$. Then we have $S = \oplus_{i} R \sigma_{i}$. We may assume that some $\sigma_{i0} = 1$. Therefore, ${}_{R}R = R \sigma_{i0}$ is a direct summand of ${}_{R}S$. Similarly, $R_{R}$ is a direct summand of $S_{R}$. So, we get the two desired conclusions.