To each representation on $\mathbb{C}^n $ corresponds a representation on $\mathbb{R}^{2n} $ So, I am supposed to show that to each representation $\rho$ of a group $G$ on $\mathbb{C}^n $ corresponds a representation $\tilde{\rho}$, of the same group $G$, on $\mathbb{R}^{2n} $. 
Additionally: Show that, if the representation $(\tilde{\rho},\mathbb{R}^{2n}) $ is irreducible, then the representation $(\rho,\mathbb{C}^{n} )$ is also irreducible. 
My thinking:
For the first part: 
Suppose we have the following map: $\phi(z):\mathbb{C}^{n}\rightarrow \mathbb{R}^{2n}, \space \phi(z)=
\begin{bmatrix}
    Re(z)   \\
    Im(z)    \\
\end{bmatrix}$. This is obviously a bijection, additionally it follows that $\phi(z_1+z_2)=\phi(z_1)+\phi(z_2)$. My idea then was to find a bijection $\eta$ between the sets $\rho(g)$ and $\tilde{\rho}(g)$, for $g \in G$ .
And this is what I came up with (I'll denote $\rho(g)$ as $\rho_g$):
$$\eta:\mathbb{R}^{2n}\rightarrow \mathbb{C}^{n}, \space \eta:\space \tilde{\rho}_g(.)\rightarrow \phi^{-1}(\tilde{\rho}_g(\phi(.)))=\rho_g(.)$$
The map $\rho_g (.)=\phi^{-1}(\tilde{\rho}_g(\phi(.)))$  is clearly $\mathbb{C}^{n} \rightarrow \mathbb{C}^{n}$, it is also a bijection and it follows that $\rho_g(z_1z_2)=\rho_g(z_1)\rho_g(z_2)$, all together meaning that $\rho_g $ is a representation of $G$ on $\mathbb{C}^{n}$. I have therefore shown that to each representation $(\tilde{\rho},\mathbb{R}^{2n})$ corresponds a representation $(\rho,\mathbb{C}^{n})$, via $\eta$.
For the second part: (proof by contradiction)
Irreducible means that for $\tilde{\rho}$ there is no (non-trivial) invariant subspace in $\mathbb{R}^{2n}$. Lets take a (non-trivial) subset $R\subset \mathbb{R}^{2n}$, so that for a subset $W \subset \mathbb{C}^{n}: \space$ $\phi(W)=R$ and $\phi^{-1}(R)=W$. We suppose that the opposite holds ($\rho$ is reducible): that $W$ is some non-trivial  invariant subset for $(\rho,\mathbb{C}^{n} )$. This means that $\rho_g(W) \subset W$ for each $g \in G$. But since $\phi^{-1}(\tilde{\rho}_g(\phi(W)))=\rho_g(W) \subset W$, it follows that $\tilde{\rho}(\phi(W))\subset R $ and using $\phi(W)=R$, we get that $\tilde{\rho}(R)\subset R$. So we found a (non-trivial) invariant subspace for $\tilde{\rho}$, but this contradicts the above statements that $\tilde{\rho}$ is irreducible, therefore $\rho$ must be irreducible.
My question:
Is my thinking correct? Could you please suggest some improvements or maybe show a different, more compact way to prove the above statements.
 A: Beware! complex $n$-dimensional representations induce real $2n$-dimensional representations but no the other way around! I believe your solution is correct, but maybe the following explanation is more compact (as you asked for). I simply avoided to work with your $\phi$ explicitly. 
The fact that $n$-dimensional representations induce complex representations follows from your correct claim that $\mathbb{C}^n \cong \mathbb{R}^{2n}$ as real vector spaces. From this, one gets that $\operatorname{GL}_{\mathbb{R}}(\mathbb{R}^{2n}) \cong \operatorname{GL}_{\mathbb{R}}(\mathbb{C}^n)$, where $\operatorname{GL}_\mathbb{R}$ denotes invertible $\mathbb{R}$-linear transformations. Since every $\mathbb{C}$-linear transformation is $\mathbb{R}$-linear, one gets a group inclusion $\operatorname{GL}_{\mathbb{C}}(\mathbb{C}^n) \hookrightarrow \operatorname{GL}_{\mathbb{R}}(\mathbb{C}^n) \cong \operatorname{GL}_{\mathbb{R}}(\mathbb{R}^{2n})$. Therefore, every complex $n$-dimensional representation $\rho\colon G \to \operatorname{GL}_{\mathbb{C}}(\mathbb{C}^n)$ induces a real $2n$-dimensional representation $\tilde{\rho}\colon G \to \operatorname{GL}_{\mathbb{C}}(\mathbb{C}^n) \hookrightarrow \operatorname{GL}_{\mathbb{R}}(\mathbb{C}^n) \cong \operatorname{GL}_{\mathbb{R}}(\mathbb{R}^{2n})$.
However, starting from a real representation $\tau\colon G \to \operatorname{GL}_{\mathbb{R}}(\mathbb{C}^n)$ nothing guarantees that it came from a complex representation, as $\mathbb{R}$-linear maps are not necessarily $\mathbb{C}$-linear. To get a complex representation out of a real one one considers an operation called complexification. The question of whether a complex representation is the complexification of a real one is very interesting and it is treated with detail in Fulton-Harris, Section 3.5.
For the question at hand, observe that if $\rho$ is reducible it admits a $G$-invariant $\mathbb{C}$-subspace. Clearly a $\mathbb{C}$-subspace is also a $\mathbb{R}$-subspace (of twice the dimension!). From our construction of $\tilde{\rho}$ (we used an inclusion!), such a subspace (under the identification $\mathbb{C}^n \cong \mathbb{R}^{2n}$) also witnesses reducibility of $\tilde{\rho}$.
