Let $C_3$ denote the cylic group of order $3$. The real representation ring $R(C_3;\mathbb{R})$ is isomorphic to $\mathbb{Z}^2$ as an abelian group. When we complexify (by tensoring representations with $\mathbb{C}$) we obtain a natural inclusion $R(C_3;\mathbb{R})\rightarrow R(C_3;\mathbb{C})\cong\mathbb{Z}^3$ where the trivial representation goes to the trivial representation and the $2$-dimensional irreducible real representation goes to the sum of the two non-trivial complex representations.
What is the analogous situation for $R(C_3;\mathbb{R})\hookrightarrow R(C_3;\mathbb{H})$ and $R(C_3;\mathbb{C})\hookrightarrow R(C_3;\mathbb{H})$. Also what is the structure of $R(C_3;\mathbb{H})$ as an abelian group?
Let $\mathbb{F}$ be either $\mathbb{C}$ or $\mathbb{H}$.
The group algebra $\mathbb{F}C_3$ has orthogonal idempotents $e_1=\frac{1}{3}(1+g+g^2)$, $e_2=\frac{1}{3}(1+\omega g+\omega^2 g^2)$ and $e_3=\frac{1}{3}(1+\omega^2 g+ \omega g^2)$, where $g$ is a generator of $C_3$ and $\omega=\frac{1}{2}(-1+i\sqrt{3})$ is a cube root of $1$.
This induces a direct sum decomposition as right $\mathbb{F}C_3$-modules $$\mathbb{F}C_3=e_1\mathbb{F}C_3\oplus e_2\mathbb{F}C_3\oplus e_3\mathbb{F}C_3.$$
The three summands are irreducible as $\mathbb{F}C_3$-modules since they are $1$-dimensional over $\mathbb{F}$.
If $\mathbb{F}=\mathbb{C}$ then they are pairwise non-isomorphic, but if $\mathbb{F}=\mathbb{H}$ then $je_2=e_3j$, so multiplication on the left by $j$ gives an isomorphism $e_2\mathbb{H}C_3\cong e_3\mathbb{H}C_3$.
So $R(C_3;\mathbb{H})\cong\mathbb{Z}^2$, generated by the classes $[e_1\mathbb{H}C_3]$ and $[e_2\mathbb{H}C_3]=[e_3\mathbb{H}C_3]$.
So the map $R(C_3;\mathbb{R})\hookrightarrow R(C_3;\mathbb{H})$ sends the classes of the two irreducible $\mathbb{R}C_3$-modules to $[e_1\mathbb{H}C_3]$ and to $2[e_2\mathbb{H}C_3]$.
The map $R(C_3;\mathbb{C})\to R(C_3;\mathbb{H})$ is not injective, as it sends the two nontrivial irreducible modules to the same place. The images of the three irreducibles are $[e_1\mathbb{H}C_3]$, $[e_2\mathbb{H}C_3]$ and $[e_2\mathbb{H}C_3]$.
In fact, as an $\mathbb{R}$-algebra, $\mathbb{H}C_3\cong\mathbb{H}\times M_2(\mathbb{C})$.