# Abstract Nonsense

## The Number of Self-Conjugate Irreps On a Finite Group of Odd Order

Point of post: In this post we use our recent works on the number of self-conjugate irrep classes of a finite group to show that for finite groups of odd order every non-trivial irrep is complex.

Motivation

We’ve done much recent work on finding different characterizations of the number of self-conjugate irrep classes of a finite group. In particular, we’ve show that if $G$ is a finite group and $\mathfrak{s}$ denotes the number of self-conjugate irreps classes of $G$ then

$\displaystyle \mathfrak{s}=\frac{1}{|G|}\sum_{g\in G}\sqrt{g}\;^2$

where $\sqrt{\text{ }}$ is the square root function for $G$. Thus, if one knew entirely the nature of the square root function on $G$ then one would know the nature of the number of self-conjugate irrep classes of $G$. Unfortunately, it is hard to say anything about the square root function on an entirely general group–in particular there is no way to calculate the square root function if $G$ is a finite group of even order. That said, as we shall see the square root function has a particularly simple description on finite groups of odd order. Pursuant to this simple nature we shall prove that if $G$ is a finite group of odd order then the only self-conjugate irrep class of $G$ is the trivial irrep class $\alpha_\text{triv}$.

The Behavior of The Square Root Function on Finite Groups off Odd Order

We begin by showing that the sets in the definition of the square root function, namely the sets $S(g)=\left\{h\in G:h^2=g\right\}$ form a partition of $G$. Giving us the fundamental relation

$\displaystyle \sum_{g\in G}\sqrt{g}=|G|$

Indeed:

Theorem: Let $G$ be a group (not necessarily finite) and define, for every $g\in G$, $S(g)$ to be the set $\left\{h\in G:h^2=g\right\}$. Then, $\left\{S(g)\right\}_{g\in G}$ forms a partition of $G$.

Proof: Clearly it’s so that $\left\{S(g)\right\}_{g\in G}$ forms a cover of $G$ since for every $g\in G$ one clearly has that $g\in S\left(g^2\right)$. Thus, it remains to prove that $S(g)\cap S(h)=\varnothing$ for $g\ne h$. But, this is relatively clear, for if $k\in S(g)\cap S(h)$ then one must have that $g=k^2=h$. From this the conclusion follows. $\blacksquare$

Corollary: Let $G$ be a finite group. Then,

$\displaystyle \sum_{g\in G}\sqrt{g}=|G|$

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Next, we show that for finite groups of odd order $G$ the square root function has a particularly simple behavior, namely:

Theorem: Let $G$ be a finite group of odd order, say $|G|=2n+1$. Then, $\sqrt{g}=1$ for every $g\in G$.

Proof: First note, using the very basic corollary of Lagrange’s theorem we know that $g^{2n+1}=e$ for every $g\in G$ and so in particular $g^{2n+2}=g$ and thus $\left(g^{n+1}\right)^2=g$ so that $g^{n+1}\in S(g)$ for every $g\in G$, in particular $\sqrt{g}\geqslant 1$ for every $g\in G$. But, by our previous theorem we have that

$\displaystyle \sum_{g\in G}\sqrt{g}=|G|$

From where it follows that $\sqrt{g}=1$ for every $g\in g$. The conclusion follows. $\blacksquare$

From this we may ascertain our main theorem. Namely:

Theorem: Let $G$ be a finite group of odd order, then every non-trivial irrep of $G$ is complex.

Proof: From our previous theorem we have that $\sqrt{g}=1$ for every $g\in G$. Thus, by our recent theorem if $\mathfrak{s}$ denotes the number of self-conjugate irrep classes of $G$, then

$\displaystyle \mathfrak{s}=\frac{1}{|G|}\sum_{g\in G}\sqrt{g}\;^2=\frac{1}{|G|}\sum_{g\in G}1=1$

But, since the trivial class $\alpha_\text{triv}$ containing the trivial irrep $\tau$ is always self-conjugate we may conclude that for every other $\alpha\in\widehat{G}$ one has that $\alpha$ is not self-conjugate, and thus complex. The conclusion follows. $\blacksquare$

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.