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The definition of equivalent representations is given by:

Definition: Let $G$ be a group, $\rho : G\rightarrow GL(V)$ and $\rho' : G\rightarrow GL(V')$ be two representations of G. We say that $\rho$ and $\rho'$ are $equivalent$ (or isomorphic) if $\exists \space T:V\rightarrow V'$ linear isomorphism such that $T{\rho_g}={\rho'_g}T\space \forall g\epsilon G$.

What I would think a more natural definition would be:

Possible different definition: Let $G$ be a group, $\rho : G\rightarrow GL(V)$ and $\rho' : G\rightarrow GL(V')$ be two representations of G. We say that $\rho$ and $\rho'$ are $equivalent$ (or isomorphic) if $\rho(G)$ is isomorphic to $\rho'(G)$.

Then we would get the nice property that:

Given $\rho : G \rightarrow GL_n(\mathbb{C})$, $\rho' : G \rightarrow GL_1(\mathbb{C})$, such that $\rho(g) = \text{Id}_n$ and $\rho'(g) = \text{Id}_1$ for all $g \in G$. Then we have $\rho \sim \rho'$.

Can one say if there would be problems with the possible different definition?

I feel it would be more natural if a definition of equivalent representations would only consider the images of the representations, and not the underlying vector spaces.

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    $\begingroup$ A representation of $G$ on $V$ can be thought of as a group action, where the bijections $V \to V$ induced by each $g \in G$ happen to be linear maps. The usual definition of equivalent representation is precisely saying that $V$ and $V’$ are isomorphic $G$-sets, and the isomorphism can be taken to be linear. So a similar question would be: why not consider two $G$-sets to be equivalent when the image of $G$ in their permutation groups are isomorphic? $\endgroup$
    – Joppy
    Commented Oct 21, 2018 at 15:06
  • $\begingroup$ @Joppy indeed: Why not consider two G-sets to be equivalent when the image of G in their permutation groups are isomorphic? $\endgroup$ Commented Oct 21, 2018 at 15:41
  • $\begingroup$ The moral reason I think is that $G$-sets or $G$-vector spaces are what we actually want to study, and hopefully having an action of $G$ helps introduce some nice symmetry into a problem. For example, in $G$-sets we get a decomposition into orbits, and in $G$-representations we get a decomposition into isotypic components. There is another reason to not use the different definition: it would mean that up to isomorphism, simple groups have precisely two representations, one trivial and one not. $\endgroup$
    – Joppy
    Commented Oct 22, 2018 at 0:46

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Firstly, see this question.

The usual notion of isomorphism implies yours: Let $\phi\colon V\to V'$ be an intertwining operator that is a linear isomorphism. Then if $\rho(g)=1$, $\rho'(g)\phi(v)=\phi(\rho(g)v)=\phi(v)$ for all $v\in V$, and $\rho'(g)=1$. The same argument the other way round shows completes the proof that $\ker\rho=\ker\rho'$, so that $\rho(G)\simeq \rho'(G)$. However, your notion is isomorphism cannot distinguish between a representation $V$ and a $V\oplus\mathbb{C}^n$ for any $n$, where $G$ acts trivially on the second factor. The major task upon meeting a new representation is to decompose it into irreducibles. For finite groups, a great too for this is character theory, which won't work for your definition.

Generally, the notion of being isomorphic for a given algebraic structure should imply that two isomorphic objects are also isomorphic as sets. Your example property shows how far your definition is from this: if you tell me you have two equivalent representations, they aren't even guaranteed to be isomorphic as vector spaces, and a representation is a vector space with extra structure.

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