Showing two groups are equal/Pontryagin duality

[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?

 

[Infrared]

$\vert H^\perp \vert = [G : H]$.

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

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

 

[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 $\vert (H^\perp)^\perp \vert = [\widehat{\widehat{G}} : H^\perp] = \frac{\vert\widehat{\widehat{G}}\vert }{\vert H^\perp\vert} = \frac{\vert G \vert}{[G: H]} = \vert H \vert$