$(\mathbb{F}_p\times\mathbb{F}_p)[F_n]\cong\mathbb{F}_p[F_n]\times\mathbb{F}_p[F_n]$ where $p$ is prime and $F_n$ is the free group of rank $n$.

Question

Let $p$ be an odd prime and $F_n$ the free group of rank $n$. I want to show the following isomorphism of group rings:

$(\mathbb{F}_p\times\mathbb{F}_p)[F_n]\cong\mathbb{F}_p[F_n]\times\mathbb{F}_p[F_n]$.

My original idea was to simply define the map $(\lambda,\,\mu)\cdot g\mapsto (\lambda\cdot g,\,\mu\cdot g)$ but a colleague said I couldn't do this. Any reason why not? And if so, how do I show this?

Answer 1

There is no need to work at this incredibly specific level of generality. It's true more generally that if $R, S$ are rings and $G$ is a group then

$$(R \times S)[G] \cong R[G] \times S[G]$$

(where the group algebra $R[G]$ over a noncommutative ring is defined so that the elements of $G$ commute with the elements of $R$). Moreover the obvious map given by extending

$$(r, s) g \mapsto (rg, sg)$$

is an isomorphism; to make this really obvious write $(r, s)$ as $(r, 0) + (0, s)$. I have no idea why your friend thinks you can't do this.