Is it true or false the following statement? If $G$ and $H$ are abelian finite groups such that $\mathbb{R}G\simeq\mathbb{R}H$, then $G\simeq H$.
I know many results when the field is $\mathbb{C}$. But I have no ideia about this case.
Any suggestions will be appreciated. Thank you.
It's false. Finite abelian groups are direct sums of the cyclic groups $C_{p^n}$ of prime power order. We have
$$\mathbb{R}[C_k] \cong \mathbb{R}[x]/(x^k - 1) \cong \mathbb{R}^r \times \mathbb{C}^s$$
where $r$ is the number of real roots of $x^k - 1$ (either $1$ or $2$ depending on whether $k$ is odd or even) and $r + 2s = k$. Moreover, we have $\mathbb{C} \otimes \mathbb{C} \cong \mathbb{C} \times \mathbb{C}$ (tensor products taken over $\mathbb{R}$). Hence we can compute, for example, that
$$\mathbb{R}[C_3 \times C_3] \cong (\mathbb{R} \times \mathbb{C}) \otimes (\mathbb{R} \times \mathbb{C}) \cong \mathbb{R} \times \mathbb{C}^4 \cong \mathbb{R}[C_9]$$
even though $C_3 \times C_3 \not\cong C_9$.
More generally, if $A$ is a finite abelian group of odd order then it has no nontrivial $1$-dimensional real representations (equivalently, no nontrivial homomorphisms into $\mathbb{R}^{\times}$, since these would have to land in the subgroup $\{ \pm 1 \}$), so $\mathbb{R}[A]$ consists of a single copy of $\mathbb{R}$ and $\frac{|A| - 1}{2}$ copies of $\mathbb{C}$, and in particular is determined by the cardinality of $A$. $C_3 \times C_3 \not \cong C_9$ is the smallest example of a pair of finite abelian groups of odd order with the same order which are not isomorphic.