Since the left regular representation is such an interesting one — in particular since it contains all the irreducible representations — we want to understand its endomorphisms. That is, we want to understand the structure of . I say that, amazingly enough, it is anti-isomorphic to the group algebra itself!
So let’s try to come up with an anti-isomorphism . Given any element , we define the map to be right-multiplication by . That is:
for every . This is a -endomorphism, since acts by multiplication on the left, and left-multiplication commutes with right-multiplication.
To see that it’s an anti-homomorphism, we must check that it’s linear and that it reverses the order of multiplication. Linearity is straightforward; as for reversing multiplication, we calculate:
Next we check that is injective by calculating its kernel. If then
so this is only possible if .
Finally we must check surjectivity. Say , and define . I say that , since
Since the two -endomorphisms are are equal on the standard basis of , they are equal. Thus, every -endomorphism of the left regular representation is of the form for some .