The rank of $R_{\mathbb{Q}}(G)$ is the number of conjugacy classes of cyclic subgroups

Choose representatives $C_1,\dots ,C_r$ of the conjugacy classes of cyclic subgroups of $G$. By [F1], the group $R_{\mathbb{Q}}(G)$ injects into ${\mathbb{Q}}^r$, so its rank is at most $r$.

For each $i$, let $U_i:={\mathrm{Ind}}_{C_i}^G{1}_{C_i}$, viewed as an element of $R_{\mathbb{Q}}(G)$. Suppose that $a_1{U}_1+\cdots +a_r{U}_r=0$ with $a_i\in \mathbb{Q}$. Reorder the representatives so that $\mathrm{\mid }{C}_1\mathrm{\mid }\ge \cdots \ge \mathrm{\mid }{C}_r\mathrm{\mid }$, and choose $i$ minimal with $a_i\mathrm{\ne }0$. Let $g_i$ be a generator of $C_i$.

If $U_j(g_i)\mathrm{\ne }0$, then [F2] and [F3] show that $g_i$ fixes some coset $x{C}_j$, so $x^{-1}{g}_{i}x\in C_j$. Therefore a conjugate of the cyclic subgroup $C_i=⟨g_i⟩$ lies in $C_j$, which implies $\mathrm{\mid }{C}_i\mathrm{\mid }\le \mathrm{\mid }{C}_j\mathrm{\mid }$. In the relation from step 1.2, every index $j<i$ has $a_j=0$ by minimality of $i$, so only $j\ge i$ can contribute. For such $j$, the ordering gives $\mathrm{\mid }{C}_j\mathrm{\mid }\le \mathrm{\mid }{C}_i\mathrm{\mid }$, hence $\mathrm{\mid }{C}_j\mathrm{\mid }=\mathrm{\mid }{C}_i\mathrm{\mid }$. A subgroup of $C_j$ with the same finite order as $C_j$ must equal $C_j$, so the conjugate of $C_i$ lying in $C_j$ is all of $C_j$. Thus $C_j$ is conjugate to $C_i$, and because $C_1,\dots ,C_r$ were chosen as distinct conjugacy-class representatives, this forces $j=i$. On the other hand, $g_i$ fixes the coset $C_i$ itself, so [F2] and [F3] give $U_i(g_i)>0$. Evaluating the relation from step 1.2 at $g_i$ therefore yields $a_i{U}_i(g_i)=0$, a contradiction. Thus the $U_i$ are linearly independent.

Step 2.1 gives $r$ linearly independent elements of $R_{\mathbb{Q}}(G)$, while step 1.1 shows that the rank is at most $r$. Hence $\mathrm{rank}{R}_{\mathbb{Q}}(G)=r$, the number of cyclic conjugacy classes.