## 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

and

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 .

**References:**

1. Simon, Barry. *Representations of Finite and Compact Groups*. Providence, RI: American Math. Soc., 1996. Print.

[...] that since if we define in the obvious way (the character of ) we may conclude that [...]

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