Usefulness of the concept of equivalent representations 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?
 A: This is just the concept of isomorphism applied to representations, i.e. $T$ is providing an isomorphism between $V$ and $V'$ which interchanges the two group
representations. So all your intuitions for the role of isomorphisms in group theory should carry over.  
Why are you having trouble seeing that properties of $\rho$ can be carried over to
$\rho'$?  As with any isomorphic situations, any property of $\rho$ that doesn't make specific reference to the names of the elements in $V$ (e.g. being irreducible or not, the number of irreducible constituents, the associated character) will carry over to $\rho'$.
A: Specific representations of groups can give us concrete information about them. Thus, much like groups themselves, we wish to classify them. To do this we need a notion of "equivalence" of representations, and this is exactly the right notion.
Without it, we are left grappling with an infinite number of representations on isomorphic vector spaces. Fortunately in nice situations, an arbitrary representation will decompose into a finite sum of a finite number of different (note we need the concept of equivalence to define a genuine notion of "diference") irreducible representations.
A similar example in group theory is that if we wish to understand a product $G \times H$, it really suffices to just understand $G$ and $H$ individually. This is what happens in the direct sum decomposition of representations.
A: 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.
