[proofplan]
By definition, a $G$-isomorphism $\varphi: V \to V'$ is a $G$-equivariant linear map that is also a vector-space isomorphism. Vector-space isomorphisms preserve dimension, since an isomorphism $\varphi$ carries a basis of $V$ to a basis of $V'$. Therefore $\dim V = \dim V'$, regardless of the $G$-action.
[/proofplan]
[step:Extract the underlying vector-space isomorphism from the $G$-isomorphism]
Suppose $\rho: G \to \operatorname{GL}(V)$ and $\rho': G \to \operatorname{GL}(V')$ are isomorphic representations. By definition, this means there exists a $G$-equivariant linear bijection
\begin{align*}
\varphi: V &\to V' \\
v &\mapsto \varphi(v)
\end{align*}
satisfying
\begin{align*}
\varphi(\rho(g)v) = \rho'(g)\varphi(v) \qquad \text{for all } g \in G,\ v \in V.
\end{align*}
In particular, $\varphi$ is a **linear bijection** $V \to V'$ — that is, an isomorphism of $\mathbb{F}$-vector spaces. The equivariance condition is irrelevant for the dimension claim and will not be used below.
[/step]
[step:Conclude $\dim V = \dim V'$ by the rank–nullity theorem]
Let $n = \dim V$ and let $\{e_1, \ldots, e_n\}$ be a basis of $V$. We claim that $\{\varphi(e_1), \ldots, \varphi(e_n)\}$ is a basis of $V'$.
**Linear independence.** Suppose $\sum_{i=1}^n c_i \varphi(e_i) = 0$ in $V'$ for scalars $c_i \in \mathbb{F}$. By linearity of $\varphi$,
\begin{align*}
\varphi\!\left(\sum_{i=1}^n c_i e_i\right) = \sum_{i=1}^n c_i \varphi(e_i) = 0.
\end{align*}
Since $\varphi$ is injective, $\sum_{i=1}^n c_i e_i = 0$ in $V$. Linear independence of the basis $\{e_i\}$ forces $c_1 = \cdots = c_n = 0$.
**Spanning.** Let $v' \in V'$. Since $\varphi$ is surjective, there exists $v \in V$ with $\varphi(v) = v'$. Expand $v = \sum_{i=1}^n a_i e_i$ with $a_i \in \mathbb{F}$. By linearity,
\begin{align*}
v' = \varphi(v) = \varphi\!\left(\sum_{i=1}^n a_i e_i\right) = \sum_{i=1}^n a_i \varphi(e_i).
\end{align*}
So $\{\varphi(e_i)\}_{i=1}^n$ spans $V'$.
Therefore $\{\varphi(e_i)\}_{i=1}^n$ is a basis of $V'$ of cardinality $n$, so $\dim V' = n = \dim V$.
[guided]
The strategy is the most direct possible: take a basis of $V$, push it forward through $\varphi$, and check that the image is still a basis. Since the image has the same cardinality as the original basis, we get $\dim V = \dim V'$.
Let $n = \dim V$ and choose any basis $\{e_1, \ldots, e_n\}$ of $V$. Set $e'_i := \varphi(e_i)$ for $i = 1, \ldots, n$. Two things to verify:
**(a) The $e'_i$ are linearly independent in $V'$.** Suppose $\sum_i c_i e'_i = 0$ for scalars $c_i \in \mathbb{F}$. By linearity of $\varphi$,
\begin{align*}
0 = \sum_i c_i e'_i = \sum_i c_i \varphi(e_i) = \varphi\!\left(\sum_i c_i e_i\right).
\end{align*}
Since $\varphi$ is **injective**, the only element of $V$ that maps to $0 \in V'$ is $0 \in V$. So $\sum_i c_i e_i = 0$. Now we use linear independence of the basis $\{e_i\}$ to conclude $c_1 = \cdots = c_n = 0$. This is where injectivity of $\varphi$ is consumed.
**(b) The $e'_i$ span $V'$.** Pick any $v' \in V'$. Since $\varphi$ is **surjective**, there exists $v \in V$ with $\varphi(v) = v'$. Expand $v$ in the basis: $v = \sum_i a_i e_i$ for some $a_i \in \mathbb{F}$. Applying $\varphi$ and using linearity:
\begin{align*}
v' = \varphi(v) = \sum_i a_i \varphi(e_i) = \sum_i a_i e'_i.
\end{align*}
This expresses $v'$ as a linear combination of the $e'_i$. This is where surjectivity of $\varphi$ is consumed.
We conclude that $\{e'_1, \ldots, e'_n\}$ is a basis of $V'$. Counting basis elements, $\dim V' = n = \dim V$, completing the proof.
A failure mode to note: if we had only assumed $\varphi$ injective, we could conclude $\dim V \leq \dim V'$; if only surjective, $\dim V \geq \dim V'$. The full bijectivity of an isomorphism is what gives the equality. The $G$-equivariance of $\varphi$ played no role at all — any vector-space isomorphism would have sufficed for the dimension conclusion.
[/guided]
[/step]
