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Let $G$ be a finite group with representations $\rho_1, \rho_2:G\rightarrow GL(V)$. According to the definition of representation isomorphisms, $\rho_1$ and $\rho_2$ are isomorphic if there exists a function $\phi:V\rightarrow V$ such that $\phi(gv)=g\phi(v)$ for all $g\in G$, $v\in V$.

Why would choosing the trivial isomorphism $\phi(v)=v$ for all $v\in V$ not show that all representations from $G$ into $V$ are isomorphic? Then $\phi(gv)=gv=g\phi(v)$? Obviously not all group representations into the same vector space are isomorphic, so what is the flaw in the reasoning here?

Thanks so much.

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You have to look more closely at what is meant with $$ \phi(gv) = g\phi(v) $$ I mean the following: On the left hand side $gv$ is shorthand for $\rho_1(g)v$, on the right, $g\phi(v) = \rho_2(g)\phi(v)$ so the equation above reads $$ \phi\bigl(\rho_1(g)v\bigr) = \rho_2(g)\phi(v), \qquad g \in G, v \in V $$ and hence $\phi = \mathrm{id}_V$ is a isomorphism from $\rho_1$ to $\rho_2$ exactly iff $$ \rho_1(g)v = \mathrm{id}_V\bigl(\rho_1(g)v\bigr) = \rho_2(g)\mathrm{id}_Vv = \rho_2(g)v, \qquad g \in G, v \in V $$ that is iff $\rho_1 = \rho_2$.

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Thanks so much! It seems so obvious now. –  William Stagner Nov 13 '12 at 7:23

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