Without using the structure theorem, how do I prove b? I struggle with the proof of injectivity. Any tips?


Let $G$ be a finite Abelian group.

(a) Prove that the group homomorphisms $\chi : G → \mathbb{C}^*$ are exactly the characters of irreducible representations of $G$.

Pointwise multiplication endows the set of irreducible characters of $G$ with the structure of a finite Abelian group. This group is denoted by $\hat{G}$. (Remark: $\hat{G}$ is also called the Pontryagin dual).

(b) Show that the map $$G \rightarrow \hat{\hat{G}}$$ $$x \mapsto (\chi \mapsto \chi(x))$$ is an isomorphism of groups.

This is what we covered in the lecture about representation theory: Representation of fine groups

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    $\begingroup$ (a) Every irreducible representation over $\mathbb{C}$ of an abelian group is one-dimensional, whence a character. $\endgroup$ – Batominovski Jul 11 '15 at 22:39
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    $\begingroup$ (b) First, you show that if $G$ is cyclic, then $\hat{G}$ is cyclic and isomorphic to $G$. Then, show that the Pontryagin dual of $G\times H$ is $\hat{G}\times\hat{H}$. This would imply that $G\cong \hat{G}$ (this result doesn't hold if $G$ is infinite, though). $\endgroup$ – Batominovski Jul 11 '15 at 22:54
  • $\begingroup$ But G is not cyclic, is it? $\endgroup$ – john Jul 11 '15 at 23:47
  • $\begingroup$ I said: IF $G$ is cyclic, then $\hat{G}$ is cyclic. $\endgroup$ – Batominovski Jul 11 '15 at 23:52

For (a) note that in an abelian group the conjugacy class of an element consists of that element only. Thus there are exactly $g=|G|$ irreducible representations and since they decompose the regular representation of $G$ (which is $g$-dimensional) the must be all $1$-dimensional.

Let $\rho$ be one of them. Then $\rho$ can be seen as a homomorphism $$ \rho:G\longrightarrow{\rm GL}_1(\Bbb C)=\Bbb C^\times $$ and also $\rho(x)={\rm tr}(\rho(x))$ for all $x\in G$, thus identifying $\rho$ with its character.

(This is also why the homomorphisms $G\rightarrow\Bbb C^\times$ of any group $G$ are called the (quasi)characters of $G$)


You want to prove that the map $$ x \mapsto (\chi \mapsto \chi(x)) $$ is injective. I assume that you have already proved that this is a homomorphism. So all you need to show is that the kernel of this map is trivial. That is, you want to show that $$ \ker (x \mapsto (\chi \mapsto \chi(x))) = \{e\} $$ where $e$ is the identity in $G$. Now the kernel is exactly all $x$ such that $$ \chi \mapsto \chi(x) $$ is the trivial map from $\hat{G}$ to $\mathbb{C}^\times$. If $x$ is in the kernel, then $\chi(x) = 1$ for all $\chi \in \hat{G}$. That is, $x$ is in the intersection of the kernels of all the characters $\chi: G\to \mathbb{C}^\times$. Hence $x$ is the identity.

  • $\begingroup$ this makes sense, but how exactly does the last argument work? That is if $x$ is in the intersection of the kernels then $x$ has to be the identity. Why is $x$ equal to the intersection and not an element of it? $\endgroup$ – john Jul 12 '15 at 0:37
  • $\begingroup$ This argument leaves off the one part of the proof that really needs justification. This is the part that my answer explains. $\endgroup$ – Seth Jul 12 '15 at 2:08
  • $\begingroup$ you're right I guess, I understand that the structure theorem is very useful to prove this, but I'm looking for a proof that doesn't rely on this theorem. This is one of many exercises that we have time to prove before the exam, but we are only allowed to use theorems we have proven in the lecture. $\endgroup$ – john Jul 13 '15 at 1:33

You can prove b) using the structure theorem for finite abelian groups. First note that a cyclic group of order $n$ maps into $\mathbb{C}^*$ by sending a generator to a primitive $n$-th root of unity.

Now to show injectivity of the map in part b), it is enough to show that the kernel is trivial, i.e. for any nonzero element $x$ of $G$ there is a character $\chi$ such that $\chi(x)\neq 1$.

Let $x\in G$ be nonzero. Then $x$ has a nonzero component in some cyclic group in the decomposition of $G$. Define a character of $G$ by mapping the chosen factor to $\mathbb{C}^*$ as explained previously, and by sending all other factors to $1$. This is gives a character $\chi$ of $G$ such that $\chi (x)\neq 1$.

  • $\begingroup$ We never had the structure theorem, only very basic representation theory. It should be possible to prove this without using the theorem. $\endgroup$ – john Jul 11 '15 at 23:47
  • $\begingroup$ @john Have you proven that characters on any subgroup extend to characters of the group? Using this result you can generate a cyclic subgroup of $x$, define a character on this cyclic subgroup as explained in my answer, and extend to a character $\chi$ such that $\chi(x)\neq 1$ as desired. $\endgroup$ – Seth Jul 13 '15 at 12:53
  • $\begingroup$ Is it the same as induced characters? No we didn't cover this either. We can of course also prove theorems and then use them. But I don't think it realistic that such big proofs are required for a subtask. The exercises are nor supposed to be easy (we get them before the exam and can use any mean to solve them) but usually there is an elementary way to do it, without the use of theorems. $\endgroup$ – john Jul 13 '15 at 13:14
  • $\begingroup$ Ok, well good luck. I don't know what results you have proven in class so I don't think I can help. $\endgroup$ – Seth Jul 13 '15 at 13:17
  • $\begingroup$ If you feel like it you can check out the lecture notes. But thanks for your help either way :) $\endgroup$ – john Jul 13 '15 at 13:56

Proof of Injectivity:

From (a) we know that the characters are homomorphic therefore $\chi(e) = e = 1$. Because the given map is also homomorphic and $e$ is maped to $(\chi \mapsto \chi(e)=1)$ the later has to be the trivial element of $\hat{\hat{G}}$. Thus the kernel of our map is $Z_{\hat{G}}:=\{g\in G:\chi(g)=1\; \forall \chi \in \hat{G}\}$. We want to show that the kernel is trivial.

Claim 1: $$\forall \chi \in \hat{G}: \sum\limits_g{\chi(g)}=\delta_{\chi,\tau}\cdot |G|$$ where $\tau(g)=1, \forall g\in G$ is the trivial representation = neutral element of $\hat{G}$

Proof: let $h \in G$ be arbitrary and fixed: $$\sum\limits_g{\chi(g)}=\sum\limits_{i=h^{-1}g}{\chi(hi)}=\sum\limits_{i}{\chi(h) \cdot \chi(i)}=\chi(h) \cdot \sum\limits_{i}{\chi(i)}=\chi(h)\cdot\sum\limits_g{\chi(g)}$$

$$ \Rightarrow \chi(h)=1\; \forall h \in G \; \text{or} \; \sum\limits_g{\chi(g)}=0$$

$$ \Rightarrow \sum\limits_g{\chi(g)}=\delta_{\chi,\tau}\cdot\sum\limits_g{\tau(g)}=\delta_{\chi,\tau}\cdot|G| $$

Claim 2: $\sum\limits_{\chi \in \hat{G}}{\chi(x)}=|G|\delta_{x \in Z_{\hat{G}}}$

Proof: let $\rho \in \hat{G}$ be arbitrary $$\sum\limits_{\chi}{\chi(x)}=\sum\limits_{\sigma=\rho^{-1}\chi}{\sigma(x)\rho(x)}=\rho(x)\sum\limits_{\sigma \in \hat{G}}{\sigma(x)}$$ $$\Rightarrow \rho(x)=1 \;\forall \rho \in \hat{G}\; or \; \sum\limits_{\chi}{\chi(x)}=0$$ $$\Rightarrow \sum\limits_{\chi}{\chi(x)}=\delta_{x \in Z_{\hat{G}}}\sum\limits_{\chi}{1} =|G|\delta_{x \in Z_{\hat{G}}}$$

Claim 3: $Z_{\hat{G}}=\{e\}$ Use Claim 1: $$ \sum\limits_{\chi}{\sum\limits_{g}{\chi(g)}}=\sum\limits_{\chi}{|G|\delta_{\chi, \tau}}=|G|$$ $$ \sum\limits_{g}{\sum\limits_{\chi}{\chi(g)}}=\sum\limits_{g}{|G|\delta_{g \in Z_{\hat{G}}}}=|G| \cdot |Z_{\hat{G}}| $$ $\Rightarrow |Z_{\hat{G}}|=1$, and as $e \in Z_{\hat{G}} \Rightarrow Z_{\hat{G}}=\{e\}$

Therefore It's proven that $\chi(g)=1, \;\forall \chi \in \hat{G} \Leftrightarrow g=e$.

  • $\begingroup$ @Seth would you say this is conclusive? $\endgroup$ – john Aug 3 '15 at 5:16

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