## The Number of Degree One Irreps

**Point of post: **In this post we prove that the number of degree one irreps of a finite group is equal to and thus, in particular, is a divisor of .

** Note:** A much better proof of this in much clearer and precise language can be found here.

*Motivation*

As of now all we know about the degrees of irreps of a finite group is that if we select one from each then . In this post we will extend our knowledge further by showing that the number of such that (recall this is just for any ) such that is equal to and in particular it divides .

*Number of Degree One Irreps*

Before we prove our main result we classify when the range of a representation is abelian. In particular:

**Theorem: ***Let be a finite group and be a representation. Then, is abelian if and only if where each is a degree one representation.*

**Proof: **Suppose first that for are degree one representations of such that . Note then that in particular each for each and . So, in particular if for every then for every one has that

and since were arbitrary it follows that is abelian as desired.

Conversely, suppose that is abelian and let for some irreps (not necessarily at this point [or obviously so] one degree). Note then that applying the previous paragraphs proof in reverse shows that the abelianess of implies the abelianess of for each . Note though that for any and any one has that is such that for any one has that and thus is an intertwinor for and itself and so by Schur’s Lemma we may conclude that for some . Since was arbitrary it follows that for every and thus every subspace of is invariant under . But, since is irreducible this is true if and only if . Since was arbitrary the conclusion follows.

What we now show is that there is an equivalence preserving bijection between irreps of and one-degree irreps of .

**Theorem: ***Let den**ote the set of all irreps on and the set of all irreps on of degree one. Then, there exists a bijection such that if and only if .*

**Proof: **Recall from our discussion of the abelinization of a group that for every group and every homomomorphism such that is abelian there exists a unique homomorphism such that . It follows then that for any representation there exists a representation such that . This is clearly a bijection, and moreover if for every then for any one has that and so . Note then that we can turn this bijection (by our last theorem) into a bijection which is also equivalence preserving. Consider then the restriction this is clearly an equivalence preserving bijection and since evidently it’s easy to see that .

**Corollary: ***The number of for which is equal to .*

**Proof: **By the previous theorem there is an equivalence preserving bijection between and one-dimensional irreps of . It follows then that and since is abelian we know that , the conclusion follows.

**References:**

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

2. Serre, Jean Pierre. *Linear Representations of Finite Groups*. New York: Springer-Verlag, 1977. Print

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