TITLE: Usefulness of the concept of equivalent representations
QUESTION [5 upvotes]: 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 $T{\rho_g}={\rho'_g}T\space \forall g\epsilon G$. 
But I don't understand why this concept is useful. If two groups $H,H'$ are isomorphic, then we can translate any algebraic property of $H$ into $H'$ via the isomorphism. But I don't see how a property of $\rho$ can be translated to similar property of $\rho'$. Nor I have seen any example in any textbook where this concept is used. Can someone explain its importance?

REPLY [9 votes]: This is true for all sorts of types of representations. Maybe it is easier to see for permutation representations.
Suppose $G$ is a group that acts on polynomials. It has an element $g$ that swaps $x$ and $y$, and leaves $z$ alone. We could write $g=(x,y)(z)$ if we wanted.
But then some jerk comes along and asks what we'd do if we needed a fourth variable. FINE. $G$ is a group that acts on polynomials. It has an element $g$ that swaps $x_1$ and $x_2$, and leaves $x_3$ alone. We could write $g=(x_1,x_2)(x_3)$ if we wanted.
Nothing important has changed really; we just changed the names of the variables.
We could go further and abbreviate $g=(1,2)(3)$. We replaced the variables with the identifying numbers. Maybe that is convenient. Saves ink (or electrons). No real change though.
Vector space representations are the same. If $G$ acts on polynomials, then I guess we could apply $g$ to $2x + 3y + 5z$ to get $2y + 3x + 5z$. So we have $\rho(g)$ a linear transformation of the vector space with basis $\{x,y,z\}$.
ARGH. I forgot about the jerk. $G$ acts on polynomials, so we could apply $G$ to $2x_1 + 3x_2 + 5x_3$ to get $2x_2 + 3x_1 + 5x_3$. So we have $\rho(g)$ a linear transformation of the vector space with basis $\{x_1,x_2,x_3\}$.
The essential properties of $\rho(g)$ and of $\rho$ are not affected by what labels we give to the basis elements of the vector space. Changing the labels on a basis is exactly and only what $T$ does. $T$ doesn't affect how $\rho(g)$ acts, it just affects where $\rho(g)$ acts.