# Abstract Nonsense

## The Dual Group

Point of post: In this post we discuss how one can turn a subset of $\widehat{G}$ naturally into a group.

Motivation

We’ve done extensive work with the set $\widehat{G}$ of all equivalency classes of irreps on a finite group $G$. A natural question that we have put off until now is if there is a natural way to make $\widehat{G}$ into a group itself. In this post we provide a partial affirmative answer to this question by showing that the set $\widehat{G_\mathfrak{L}}$ of all linear representations (representations of degree one) forms a group under a natural product.

Linear Representations and Their Tensor Products

$\text{ }$

Let $G$ be a finite group an irrep $\rho$ of $G$ is said to be linear if $\deg\rho=1$. We call an equivalency class $\alpha\in\widehat{G}$ linear if $d_\alpha=1$ (or equivalently if every one of its members is a linear irrep). If $\lambda$ is a linear irrep then $\chi_\lambda$ is called a linear character of $G$. We denote the set of all linear $\alpha\in\widehat{G}$ by $\widehat{G_\mathfrak{L}}$.

$\text{ }$

What we first show is that if $\rho$ is an irrep and $\lambda$ a linear irrep then the tensor product $\lambda\otimes\rho$ is an irrep. Indeed:

$\text{ }$

Theorem: Let $G$ be a finite group. Then, if $\lambda$ is a linear irrep of $G$ and $\rho$ an irrep then $\lambda\otimes\rho$ is an irrep of $G$ with $\deg(\lambda\otimes\rho)=\deg\rho$.

Proof: To see that $\lambda\otimes\rho$ is irreducible we note that

\displaystyle \begin{aligned}\left\langle\chi_{ \lambda\otimes\rho},\chi_{\lambda\otimes\rho}\right\rangle &= \frac{1}{|G|}\sum_{g\in G}\left|\chi_\lambda(g)\chi_\rho(g)\right|^2\\ &=\frac{1}{|G|}\sum_{g\in G}\left|\chi_\lambda(g)\right|^2\left|\chi_\rho(g)\right|^2\\ &= \frac{1}{|G|}\sum_{g\in G}\left|\chi_\rho(g)\right|^2\\ &=\left\langle\chi_\rho,\chi_\rho\right\rangle\\ &= 1\end{aligned}

(where we’ve used the simple fact that for any linear representation $\lambda$ one has that $\left|\chi_\lambda(g)\right|=1$ for every $g\in G$) from where the conclusion follows from our previous characterization of irreducibility. To see that $\deg(\lambda\otimes\rho)=\deg\rho$ we merely note that

$\text{ }$

$\deg(\lambda\otimes\rho)=\chi_{\lambda\otimes\rho}(e)=\chi_\lambda(e)\chi_\rho(e)=\chi_\rho(e)=\deg\rho$.

$\blacksquare$

$\text{ }$

$\text{ }$

Now, it’s clear from this that for any two linear irreps $\lambda,\mu$ their tensor product $\lambda\otimes\mu$ is also a linear irrep, and since it’s evident that if $\lambda\cong\lambda'$ and $\mu\cong\mu'$ that $\lambda\otimes\mu\cong\lambda'\otimes\mu'$ we may unambiguously consider the product of $\alpha,\beta\in\widehat{G_\mathfrak{L}}$, denoted $\alpha\otimes \beta$ to be the equivalency class $[\lambda\otimes\mu]$ for any $\lambda\in\alpha$ and $\mu\in\beta$.

$\text{ }$

It’s evident that for any three representations $\rho,\psi,\nu$ that $\rho\otimes(\psi\otimes \nu)\cong(\rho\otimes \psi)\otimes \nu$ so that if $\alpha,\beta,\gamma$ are the equivalency classes of $\rho,\psi,$ and $\nu$ respectively then $\alpha\otimes(\beta\otimes\gamma)=(\alpha\otimes\beta)\otimes\gamma$ so that the map $\otimes:\widehat{G_\mathfrak{L}}\times\widehat{G_\mathfrak{L}}\to\widehat{G_\mathfrak{L}}$ is associative. Moreover, it’s clear that if $\alpha_{\text{triv}}$ denotes the equivalence class of the trivial character $\tau:G\to\mathbb{C}^\times:g\mapsto 1$ then $\alpha_\text{triv}$ is a two-sided identity for $\otimes$. Indeed, recall that two irreps are equivalent if and only if the admit the same character and for any $\alpha\in\widehat{G_\mathfrak{L}}$ and $\lambda\in\alpha$ one has that

$\text{ }$

$\chi_{\tau\otimes\lambda}(g)=\chi_\tau(g)\chi_\lambda(g)=\chi_\lambda(g)=\chi_\lambda(g)\chi_\tau(g)=\chi_{\lambda\otimes\tau}(g)$

for every $g\in G$ so that $\tau\otimes\lambda\cong\lambda\cong\lambda\otimes\tau$ or, said equivalently, $\alpha_{\text{triv}}\otimes\alpha=\alpha=\alpha\otimes\alpha_{\text{triv}}$. Lastly, we claim that if $\alpha\otimes\beta=\alpha\otimes\gamma$ then $\beta=\gamma$. Indeed,

\displaystyle \begin{aligned}\delta_{\beta,\gamma} &= \frac{1}{|G|}\sum_{g\in G}\chi^{(\beta)}(g)\overline{\chi^{(\gamma)}(g)}\\ &= \frac{1}{|G|}\sum_{g\in G}\chi^{(\beta)}(g)\overline{\chi^{(\gamma)}(g)}\left|\chi^{(\alpha)}(g)\right|^2\\ &= \frac{1}{|G|}\sum_{g\in G}\chi^{(\alpha)}(g)\chi^{(\beta)}(g)\overline{\chi^{(\alpha)}(g)\chi^{(\gamma)}(g)}\\ &= \frac{1}{|G|}\sum_{g\in G}\chi^{(\alpha\otimes\beta)}(g)\overline{\chi^{(\beta\otimes\gamma)}(g)}\\ &= \left\langle \chi^{(\alpha\otimes\beta)},\chi^{(\alpha\otimes\gamma)}\right\rangle\\ &=1\end{aligned}

Thus, $\widehat{G_\mathfrak{L}}$ is a finite set which has the associative binary operation $\otimes$ with the two-sided identity $\alpha_\text{triv}$ and which satisfies the left and right cancellative property. There’s a name for such a structure: a group. Moreover, it’s clear that $\widehat{G_\mathfrak{L}}$ is commutative since for any any two representations $\rho$ and $\psi$ it’s clearly true that $\rho\otimes\psi\cong\psi\otimes \rho$. Also, since for any $\alpha\in\widehat{G_\mathfrak{L}}$ and any $\lambda\in\alpha$ we have that $1=\left|\chi_\lambda(g)\right|^2=\chi_\lambda (g)\overline{\chi_\lambda(g)}$ it’s clear that $\chi_{\lambda\otimes\text{Conj}_\lambda^J}=\chi^{\text{triv}}$ for any complex conjugate $J$ on $\lambda's$ representation space, and so $\alpha^{-1}=\overline{\alpha}$. Summing this all up we have:

$\text{ }$

Theorem: Let $G$ be a finite group, then $\left(\widehat{G_\mathfrak{L}},\otimes\right)$ is an abelian group with identity $\alpha_\text{triv}$ and $\alpha^{-1}=\overline{\alpha}$.

$\text{ }$

We call the group $\widehat{G_\mathfrak{L}}$ the dual group of $G$.

$\text{ }$

As a last note, we know that $\left|\widehat{G_\mathfrak{L}}\right|=\left|G^\text{ab}\right|=\#\left(G/[G,G]\right)$ where $G^\text{ab}$ is the abelianization of $G$ and $[G,G]$ is the commutator subgroup.

$\text{ }$

$\text{ }$

References:

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

April 12, 2011 -

1. […] our last post we discussed how to naturally turn a subset of into a group, namely the subset of linear […]

Pingback by Representation Theory: Another Way of Viewing the Dual Group « Abstract Nonsense | April 12, 2011 | Reply

2. […] it’s difficult to deal directly with the dual group even with our alternate characterization of it. In this post we show that the lay of the land […]

Pingback by Representation Theory: Dual Group of a Cyclic Group « Abstract Nonsense | April 13, 2011 | Reply

3. […] We recall that the dual group  has order equal the the abelianization . But, by inspection we see that and so and so […]

Pingback by University of Maryland College Park Qualifying Exams (Group Theory and Representation Theory) (August-2003) « Abstract Nonsense | May 1, 2011 | Reply

4. […] and to be honest I was always a little hazy on the end of it. A while ago when I was thinking about dual groups and trying to develop stuff independent of my books I realized that there was a ‘much […]

Pingback by Clearer Proof for the Number of Degree One Irrep Classes « Abstract Nonsense | May 6, 2011 | Reply