# Showing two groups are equal/Pontryagin duality

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:

Infrared
Gold Member
##\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
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