In the book Algebra by Cohn, he gives the following definitions
Two representations $\rho,\sigma$ of a group are said to be equivalent if they have the same degree and there exists an invertible matrix $P$ such that for any $x\in G$
$$
\sigma=P^{-1}\rho(x)P.
$$
Given two $G$-modules $U$ and $V$, a $G$-isomorphism from $U$ to $V$ is a bijective linear mapping $f :U\rightarrow V$ such that
$$
f(ux)=(fu)x,\qquad \text{for all }u\in U, x\in G
$$
Later he writes
If two $G$-modules afford the same representation, they must be isomorphic. For take the modules to be $V, W$ with bases $v=v_1,…,v_d)^T, w=(w_1,…,w_d)^T$ and let
$$
v_i x=\sum_j\rho_{ij}(x)v_j,\qquad w_i x=\sum_j\rho_{ij}(x)w_j.
$$
Then the mapping $\sum\alpha_iv_i\mapsto\sum\alpha_iw_i$ is a $G$-isomorphism between $V$ and $W$. It follows that isomorphic modules afford equivalent representations, for by changing the bases in one of the modules we can make the representations equal.
The last part in italic is what I do not understand. To me it seems like he says "Two $G$-modules that afford the same representation are isomorphic, therefore isomorphic modules afford equivalent representations", which is not a correct mathematical argument (to be clear, it seems like he shows $\Longrightarrow$ and uses this to say that $\Longleftarrow$, which obviously isn't necessarily the case). Can someone help me understand what he means?
Best Answer
I agree - as written, it seems like the author explains the implication in one direction and then claims that this proves the implication in the other direction too.
To prove the implication in the other direction, we can argue as follows. Suppose that $U$ and $V$ be two isomorphic $G$-modules, with bases $\mathcal B_U = \{ u_1, \dots, u_d \}$ and $\mathcal B_V = \{ v_1, \dots, v_d \}$.
Given any vector $u \in U$, let $u_{\mathcal B_U} \in \mathbb C^d$ denote the components of $u$ with respect to the basis $\mathcal B_U$. Similarly, given any vector $v \in V$, let $v_{\mathcal B_V} \in \mathbb C^d$ denote the components of $v$ with respect to the basis $\mathcal B_V$.
For any $x \in G$, let $\rho_{\mathcal B_U}(x) \in \text{GL}(d, \mathbb C)$ be the matrix representing the action of the group element $x$ on $U$ with respect to the basis $\mathcal B_U$, and let $\sigma_{\mathcal B_V}(x) \in \text{GL}(d, \mathbb C)$ be the matrix representing the action of $x$ on $V$ with respect to the basis $\mathcal B_V$. Thus $ ux = \rho_{\mathcal B_U}(x) \ u_{\mathcal B_U}$ for any $x \in X$ and $u \in U$, and $vx = \sigma_{\mathcal B_V}(x) \ v_{\mathcal B_V}$ for any $x \in X $ and $v \in V$.
Suppose $f: U \to V$ is a $G$-module isomorphism between $U$ and $V$, and let $f_{\mathcal B_U, \mathcal B_V} \in \text{GL}(d, \mathbb C)$ be the matrix for $f$ with respect to the bases $\mathcal B_U$ and $\mathcal B_V$. Then $(f(u))_{\mathcal B_V} = f_{\mathcal B_U, \mathcal B_V} \ u_{B_U}$ for any $u \in U$.
Since $f: U \to V$ is a $G$-module isomorphism, we have $f(ux) = f(u)x$ for any $u \in U$ and for any $x \in X$. Thus $$ f_{\mathcal B_U, \mathcal B_V} \ \rho_{\mathcal B_U}(x) \ u_{\mathcal B_U} = (f(ux))_{\mathcal B_V} = (f(u)x)_{\mathcal B_V} = \sigma_{\mathcal B_V}(x) \ \ f_{\mathcal B_U, \mathcal B_V} \ u_{\mathcal B_U},$$ and since this holds for arbitrary $u \in U$, we have the matrix equation $$ f_{\mathcal B_U, \mathcal B_V} \ \rho_{B_U}(x) = \sigma_{\mathcal B_V}(x) \ f_{\mathcal B_U, \mathcal B_V}.$$
Thus $$ \rho_{\mathcal B_U}(x) = (f_{\mathcal B_U, \mathcal B_V})^{-1} \ \sigma_{\mathcal B_V}(x) \ f_{\mathcal B_U, \mathcal B_V},$$ which shows that $\rho$ and $\sigma$ are equivalent representations.
Finally, you may be wondering what is the link between my argument on the sentence in italics from your book.
To see the link, observe that $(f_{\mathcal B_U, \mathcal B_V})^{-1} \ \sigma_{\mathcal B_V}(x) \ f_{\mathcal B_U, \mathcal B_V}$ is the matrix representation of the action of $x$ on $V$ with respect to the basis $f(\mathcal B_U) := \{ f(u_1), \dots, f(u_d) \}$.