I am learning the group algebra. A way to study the irreducible representations of a finite group is to use the idempotent operators in the group algebra.
Suppose $e$ and $e'$ are the primitive idempotent operators associated with two left ideals $I$ and $I'$. Then, the right multiplication by $exe'$ ($x$ an arbitrary element in the group algebra) defines a linear mapping from $I$ to $I'$, which has the great property of commuting with the left action of the group elements on $I$ and $I'$.
The problem is, $xe'$ has also such properties.
So, is the $e$ superfluous in $exe'$?
I think by "associated with" you mean that $Re=I$ and $Re'=I'$.
It is true that the map induced by $xe'$ On $Re$ is the same as the map induced by $exe'$ : $(re)xe'=(re)exe'$ for all $r\in R$.
However, $exe'$ has the additional property that it is zero on the complement $R(1-e)$. For another thing, $eRe'\cong Hom(_RRe, _RRe')$, so it makes sense to deliberately choose maps according to this isomorphism.. Both ways the choice is much 'cleaner.'