This may be totally trivial or wrong. I am just posting this because I am sick and tired of trying to understand this myself and I am sure someone out here can just answer it out of his head in 2 minutes.

Let $G$ be a finite group, and $k$ an algebraically closed field whose characteristic is not a divisor of $\left|G\right|$. The Maschke theorem in its standard form states that $$ \displaystyle k\left[G\right]\cong\bigoplus_{V\text{ irreducible }k\left[G\right]\text{-module}}\mathrm{End}V $$ as $k$-algebras. This quickly yields that $$ \displaystyle k\left[G\right]\cong\bigoplus_{V\text{ irreducible }k\left[G\right]\text{-module}}\mathrm{End}V $$ as left $k\left[G\right]$-modules, where the $k\left[G\right]$-module structure on $\mathrm{End}V$ is given by $\left(gF\right)\left(v\right)=gF\left(v\right)$ for every $g\in k\left[G\right]$, $f\in\mathrm{End}V$ and $v\in V$.

Now, I've overheard that there exists a stronger form of this, stating that $$ \displaystyle k\left[G\right]\cong\bigoplus_{V\text{ irreducible }k\left[G\right]\text{-module}}\mathrm{End}V $$ as left $k\left[G\times G\right]$-modules, where the $k\left[G\times G\right]$-module structure on $k\left[G\right]$ is defined by $\left(g,h\right)u=guh^{-1}$ for any $g\in G$, $h\in G$ and $u\in k\left[G\right]$, and the $k\left[G\times G\right]$-module structure on $\mathrm{End}V$ is defined by $\left(g,h\right)F=gFh^{-1}$ for any $g\in G$, $h\in G$ and $F\in \mathrm{End}V$. This would follow from $$ \displaystyle k\left[G\right]\cong\bigoplus_{V\text{ irreducible }k\left[G\right]\text{-module}}\mathrm{End}V $$ as $k$-Hopf algebras, where the Hopf algebra structure on $k\left[G\right]$ is the standard one ($S\left(g\right)=g^{-1}$ for every $g\in G$), and the Hopf algebra structure on $\mathrm{End}V$ is the standard one as well.

[**EDIT:** as the comments explained, there is no such thing as a standard Hopf algebra structure on $\mathrm{End}V$, and so "$k$-Hopf algebras" should be "$k$-algebras".]

Is this correct, and how can this be proven?

knowthat what is being bracketed is not going to be taller than normal (and in fact, you do not want in general for the brackets to grow even if the inside is slightly taller than ordinary---think of an exponent) – Mariano Suárez-Alvarez♦ Jan 12 '10 at 23:18