Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I am trying to learn about characters of finite abelian groups. A character is a homomorphism from a finite abelian group $G$ into the multiplicative group of complex numbers of absolute value 1. In my textbook (Finite Fields by Lidl and Niederreiter) there is the following question to which I am stuck:

Let $H$ be a subgroup of the finite abelian group $G$. Prove that the annihilator $A$ of H in $\widehat{G}$ (where $\widehat{G}$ is the group of characters of $G$) is isomorphic to $G/H$ and that $\widehat{G}/A$ is isomorphic to $H$.

This looks like something where the 1st isomorphism theorem for groups could be used, but I don't see how. Any ideas would be greatly appreciated.

share|cite|improve this question
up vote 0 down vote accepted

First question

Let $ {\text{Ann}_{G}}(H) $ denote the annihilator of $ H $ in $ G $, i.e., $$ {\text{Ann}_{G}}(H) = \left\{ \phi \in \widehat{G} ~ \middle| ~ \forall h \in H: ~ \phi(h) = 1_{\mathbb{C}} \right\}. $$ Then $ {\text{Ann}_{G}}(H) \cong \widehat{G / H} $. In order to prove this, let $ q: G \to G / H $ denote the obvious quotient group homomorphism, and define a group homomorphism $ \Phi: \widehat{G / H} \to {\text{Ann}_{G}}(H) $ by $$ \forall \phi \in \widehat{G / H}: \quad \Phi(\phi) = \phi \circ q. $$

$ \Phi $ is injective: Suppose that $ \phi \in \widehat{G / H} $ and $ \phi \circ q = \mathbf{1}_{G} $. Then $ \phi(g + H) = 1_{\mathbb{C}} $ for all $ g \in G $, so $ \phi = \mathbf{1}_{G / H} $.

$ \Phi $ is surjective: Let $ \phi \in {\text{Ann}_{G}}(H) $. We can define a map $ \dot{\phi}: G / H \to \mathbb{C} $ by $$ \forall g \in G: \quad \dot{\phi}(g + H) \stackrel{\text{df}}{=} \phi(g). $$ This is clearly a well-defined map and is a character on $ G / H $. As $ \phi = \dot{\phi} \circ q $, we are done.

Note: Up to this point, all of our arguments are valid for an arbitrary locally compact Hausdorff abelian group $ G $ with $ H $ a closed subgroup and all maps involved continuous.

We now turn to the special case when $ G $ is finite and abelian with the discrete topology.

Question. How should we use the assumption that $ G $ is finite and abelian?

It turns out that any finite and abelian group is isomorphic to its own dual. By assumption, $ G $ is finite and abelian, so $ G / H $ is also finite and abelian. Hence, $ G / H \cong \widehat{G / H} $, which yields $$ {\text{Ann}_{G}}(H) \cong G / H $$ as desired. Note, however, that the isomorphism between $ G / H $ and $ \widehat{G / H} $ is not natural.

Second question

Let us first show that $ \widehat{G} / {\text{Ann}_{G}}(H) \cong \widehat{H} $. By the answer to the first question, we have $$ (\spadesuit) \qquad \left( \widehat{G} / {\text{Ann}_{G}}(H) \right)^{\land} \cong {\text{Ann}_{\widehat{G}}}({\text{Ann}_{G}}(H)). $$

Claim: $ {\text{Ann}_{\widehat{G}}}({\text{Ann}_{G}}(H)) \cong H $.

Proof of Claim

Observe that \begin{align} {\text{Ann}_{\widehat{G}}}({\text{Ann}_{G}}(H)) & = \left\{ \Psi \in \widehat{\widehat{G}} ~ \middle| ~ \forall \phi \in {\text{Ann}_{G}}(H): ~ \Psi(\phi) = 1_{\mathbb{C}} \right\} \\ & \cong \{ g \in G \mid \forall \phi \in {\text{Ann}_{G}}(H): ~ \phi(g) = 1_{\mathbb{C}} \} \quad (\text{By Pontryagin Duality.}) \\ & \supseteq H. \quad (\text{By the definition of $ {\text{Ann}_{G}}(H) $.}) \end{align} It then remains to prove that we actually have equality in the last line. As $ {\text{Ann}_{G}}(H) \cong \widehat{G / H} $, we get $ \left( {\text{Ann}_{G}}(H) \right)^{\land} \cong G / H $ by Pontryagin Duality. The isomorphism is explicitly implemented by the map $ \Theta: G / H \to \left( {\text{Ann}_{G}}(H) \right)^{\land} $ defined by $$ \forall g \in G, ~ \forall \phi \in {\text{Ann}_{G}}(H): \quad [\Theta(g + H)](\phi) \stackrel{\text{df}}{=} \phi(g). $$ If $ g \notin H $, then $ g + H \neq e_{G / H} $, so there exists a $ \phi \in {\text{Ann}_{G}}(H) $ such that $$ \phi(g) = [\Theta(g + H)](\phi) \neq 1_{\mathbb{C}}. $$ This readily implies that $$ \{ g \in G \mid \forall \phi \in {\text{Ann}_{G}}(H): ~ \phi(g) = 1_{\mathbb{C}} \} = H $$ as desired. $ \quad \blacksquare $

It now follows from $ (\spadesuit) $ that $ \left( \widehat{G} / {\text{Ann}_{G}}(H) \right)^{\land} \cong H $. By Pontryagin Duality yet again, $$ \widehat{G} / {\text{Ann}_{G}}(H) \cong \widehat{H}. $$ Finally, as $ H $ is finite and abelian, we obtain $ \widehat{H} \cong H $, which concludes the argument.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.