Question about the group algebra

Question

Let $G$ be a group and let $k[G]$ be the group algebra of $G$. If we let $H$ be a subgroup of $G$ and $T=\{t_1,...,t_n\}$ a right transversal for $H$, then we can write $$ G=\bigcup_{i=1}^nHt_i. $$ My question is, does this imply that we can write the group algebra as $$ k[G]=\bigoplus_{i=1}^nk[Ht_i], $$ where $k[Ht_i]$ is the group algebra of the subgroup $Ht_i$?

Answer 1

As pointed out in the comments, cosets $Ht$ are not subgroups (unless $t\in H$, of course), so we wouldn't write something like $k[Ht]$ in the sense of group algebras. The notation $k[H]t$ makes sense, though, as the image of $k[H]$ under right-multiplication-by-$t$, i.e. $\{at\mid a\in k[H]\}$. We could also use the notation $kHt$ and interpret it as the free vector space spanned by $Ht$ within $k[G]$ (assuming we're treating $G$ itself as a $k$-vector space basis for the group algebra $k[G]$). Thus, for instance, if $H=\{h_1,h_2,h_3,\cdots\}$ then the elements of $k[H]t$ or $kHt$ are linear combinations of $h_1t,h_2t,h_3t,\cdots$.

And yes, the fact we can partition $G$ into cosets does mean we can write

$$ k[G]=\bigoplus_{t\in T} k[H]t $$

as $k$-vector spaces, and indeed as left $k[H]$-modules (aka $H$-reps).