Representation of an abelian groups

Question

How do I prove $(b)$? I struggle with the proof of injectivity. Any tips?

$Problem$: 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 $$\mathcal{H} :G \rightarrow \hat{\hat{G}}$$ $$a \mapsto (\chi \mapsto \chi(a))$$ is an isomorphism of groups.

Answer 1

Here is a hands on proof of injectivity of $\mathcal{H}$: First note that for any finite abelian group $G$ and fixed $\chi\in \hat G$ we have

$$ \sum_{a\in G} \chi(a) = |G| \delta_{\chi,1}, \tag{$\star$} $$

where $1\in \hat G$ is the trivial representation. Indeed let $b\in G$ then

$$\chi(b) \sum_{a} \chi(a) = \sum_a \chi(ba) = \sum_c \chi(c),$$

which implies that either $\sum_{a} \chi(a)= 0$ or $\chi = 1$.

Now let us compute

$$ \sum\limits_{\chi}{\sum\limits_{a }{\chi(a)}}\overset{(\star)}{=}\sum\limits_{\chi}{|G|\delta_{\chi,1}}=|G|$$

since the group is finite we can interchange the sums

$$ {\sum\limits_{a } \sum\limits_{\chi} {\chi(a)}} = \sum\limits_{a } \sum\limits_{\chi} \mathcal{H}(a)(\chi) \overset{(\star)}{=}\sum\limits_{a}{|\hat G|\delta_{\mathcal{H}(a),\hat 1}} = |G| |\ker{\mathcal{H}}| $$

where we apply $(\star)$ to the finite abelian group $\hat G$, $\mathcal{H}(a)$ takes the roll of the fixed $\chi$ in $(\star)$ and $\hat 1 \in \hat {\hat G}$ is the trivial representation of $\hat G$.

Answer 2

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 $\chi (x)\neq 1$.

Answer 3

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 $G$)

Answer 4

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.