# Showing two groups are equal/Pontryagin duality

• fishturtle1
In summary: G## is abelian. Therefore, we have shown ##(H^\perp)^\perp## has the same number of elements as ##H##, proving ##(H^\perp)^\perp = H##. In summary, we are trying to show that ##H = (H^\perp)^\perp## and we have proven this by showing that ##(H^\perp)^\perp## has the same number of elements as ##H##, which is ##\vert H \vert##. This is done using the fact that ##G## is abelian and the definitions given above.
fishturtle1
Homework Statement
Let ##G## be a finite group and ##H \subset G## be a subgroup of ##G##.

a) Viewing ##H^{\perp\perp} = (H\perp)^\perp)## in G using Pontryagin duality, show ##H^{\perp\perp} = H##. (Hint: The inclusion in one direction should be easy. Count sizes for the other inclusion).
Relevant Equations
##\widehat{G} = \lbrace \chi : G \rightarrow S^1 : \chi(gh) = \chi(g)\chi(h), \forall g, h \in G \rbrace##.

##H^{\perp} = \lbrace \chi \in \widehat{G} : \chi = 1 \text{on H} \rbrace##

Pontryagin duality says ##G = \widehat{\widehat{G}}## and we have the isomorphism ##f: G \rightarrow \widehat{\widehat{G}}## defined by ##g \mapsto \sigma_g## where ##\sigma_g : \widehat{G} \rightarrow S^1## is a homomorphism defined by ##\sigma_g(\chi) = \chi(g)## for all ##\chi \in \widehat{G}##. (this is how I understood it from the notes, but maybe I misunderstood something?).

Also, in the previous exercise we showed ##\vert H^\perp \vert = [G : H]##.
I am confused because ##H## is a subgroup of ##G## and ##H^{\perp\perp}## is a set of homomorphisms. Are we trying to show ##f(H) = H^{\perp\perp}## where ##f## is the isomorphism defined in the definitions above?

Proof: We want to show ##H = (H^\perp)^\perp##.

##(\subset):## Let ##h \in H##. Consider ##\sigma_h## (as defined in the above definitions). Then, ##\sigma_h(\chi) = \chi(h) = 1## for all ##\chi \in H^{\perp}##. So, ##\sigma_h \in (H^\perp)^\perp##. So, ##f(H) \subset (H^\perp)^\perp##.

##(\supset):## We want to show there's at most ##\vert H \vert## elements in ##(H^\perp)^\perp##. Let ##\psi_1, \psi_2 \in (H^\perp)^\perp## and suppose ##\psi_1 \neq \psi_2##. We have ##\psi_1\vert_{\widehat{H}}, \psi_2\vert_{\widehat{H}} \in \widehat{\widehat{H}}##. Since ##\psi_1\neq\psi_2##, there exists some ##\chi \in \widehat{H}## such that ##\psi_1(\chi) \neq \psi_2(\chi)##. We conclude ##\psi_1 \neq \psi_2## on ##\widehat{H}## and so ##(H^{\perp})^\perp## has at most ##\vert \widehat{\widehat{H}} \vert = \vert H \vert## elements.

Is what I've written correct so far?

Last edited:
fishturtle1 said:
##\vert H^\perp \vert = [G : H]##.

This should tell you how to find ##|(H^\perp)^\perp|.##

Edit: I think you forgot to assume that ##G## should be abelian.

Last edited:
fishturtle1
Infrared said:
This should tell you how to find ##|(H^\perp)^\perp|.##
We want to show ##\vert (H^\perp)^\perp \vert = \vert H \vert##. We have ##\vert (H^\perp)^\perp \vert = [\widehat{\widehat{G}} : H^\perp] = \frac{\vert\widehat{\widehat{G}}\vert }{\vert H^\perp\vert} = \frac{\vert G \vert}{[G: H]} = \vert H \vert##

Infrared

## 1. How can I show that two groups are equal?

In order to show that two groups are equal, you must first prove that they have the same elements and operations. This can be done by showing that every element in one group has a corresponding element in the other group, and that the operations on these elements produce the same results in both groups.

## 2. What is Pontryagin duality?

Pontryagin duality is a mathematical concept that relates two groups, the dual group and the Pontryagin dual group, through a process called duality. It states that for any locally compact abelian group, its dual group is isomorphic to its Pontryagin dual group, which is a group of characters (continuous homomorphisms) on the original group.

## 3. How is Pontryagin duality used in mathematics?

Pontryagin duality is used in a variety of mathematical fields, including harmonic analysis, number theory, and representation theory. It allows for the study of a group through its dual group, which can provide a different perspective and lead to new insights and theorems.

## 4. What are the applications of Pontryagin duality?

Pontryagin duality has many applications in mathematics, including in the study of Fourier analysis, topological groups, and the theory of algebraic curves. It is also used in physics, particularly in quantum mechanics, where it relates the position and momentum operators through duality.

## 5. Are there any limitations to Pontryagin duality?

While Pontryagin duality is a powerful tool in mathematics, it is limited to locally compact abelian groups. This means that it cannot be applied to non-abelian groups or groups that are not locally compact. Additionally, the duality may not always provide a complete understanding of the group, and other techniques may be needed to fully characterize it.

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