## The Square Root Function and its Relation to Irreducible Characters

**Point of post: **In this post we describe what can best be verbally described as “the number of square roots” function for a group and a way which it relates to the irreducible characters of the group.

**Motivation**

Recall that in our last post that we found an interesting property involving the characters: namely, we characterized real, complex, and quaternionic irreps in terms of their character. In this characterization a sum come up whose summand had the form . That said, since the major theorems we have thus far developed involve summands of the form it would, of course, be preferable to change the summand in our characterization of real, complex, and quaternionic irreps into a summand involving . The way we can do this is clear, namely for each we define the ‘square root’ of , denoted , to be equal to . Then, with this it’s clear that our characterization can be rewritten as a sum with summand . It turns out though that the interplay goes much farther than this, to the point where we can actually express entirely in terms of irreducible characters…and thus make it possible to compute from a groups character table.

*The Square Root Function and its Relation to Irreducible Characters*

Let be a finite group and define the map by . We call this the *square root function on .* Then, our previous characterization of real,complex, and quaternionic irreps can be phrased in terms of . Indeed, let be an irrep of the finite group . Then,

Where is and if is real, complex, and quaternionic respectively. Note that the notation is appropriate (in the sense that it only depends on ), since being a real, complex, or quaternionic irrep is invariant under equivalence.

To derive our main result we need to know very little about itself. In fact, we only need the following basic theorem:

Theorem:Let be defined as above. Then, is a class function.

**Proof: **Let be arbitrary. Define then by . This mapping is well defined (in the sense that the image of the map really lies in the codomain) since if then . Evidently then is injective (being the restriction of the inner automorphism ) and thus it suffices to prove that is surjective. To see this suppose that is an element of the codomain, then and so but, of course and so is an element of the codomain. But, a quick check shows then that . Since was arbitrary the conclusion follows. Thus, since is a bijection we may conclude that

Since were arbitrary we may conclude that as required.

With this theorem in mind we are now well-equipped to prove the main result of this post. Namely:

Theorem:Let be a finite group and . Then,

* *

**Proof: **Citing once again the formula we may conjugate both sides to obtain

Then, multiplying both sides by gives

Then, summing over and performing minimal manipulations to the left-hand gives

But, using the second orthogonality relation and denoting the conjugacy class of by we may rewrite this as

Which, of course may be rewritten as

But, by our previous theorem we know that is a class function and so the left-hand side reduces to

The conclusion follows.

**References:**

1. Isaacs, I. Martin. *Character Theory of Finite Groups*. New York: Academic, 1976. Print.

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

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