Another Way of Looking at Induced Representations (Pt. II)
Point of Post: This post is a continuation of this one.
Our Two Formulations Are Equivalent
What we’d now like to show is that and are actually equivalent. Indeed:
Theorem: Let be a finite group, , and . Then, if is our first definition of induced representation and the one developed in this post, then .
Proof: We define
by the rule
where is the unique representation of an element of as the product of something in and something in . This is well-defined, in the sense that , since if and then
In fact, we’ve already seen this map in a sense before when finding and have seen that it is a bijection and has inverse
What we next claim is that for every one has . To do this we first note that
So that by definition (using the obvious fact that for fixed , is a bijection )
But, from previous work with we know that
so that may be restated as
And, since was arbitrary we may appeal to a previous theorem.
Remark: Note that this proves as well the implied fact that, up to equivalence, the construction of is independent of choice of transveral .
1. Simon, Barry. Representations of Finite and Compact Groups. Providence, RI: American Math. Soc., 1996. Print.