Problem 5.2.14 (Dual group)

Proof. Let $G$ be a finite group $Ĝ=\mathrm{Hom}(G,{ℂ}^{\ast })$ be the set of all homomorphisms from $G$ into the multiplicative group ${ℂ}^{\ast }$. Let $\chi \in Ĝ$. Since every element $x\in G$ has a finite order $r$, then $x^r=1$ implies $\chi {(x)}^r=1$, so $|\chi (x)|=1$ and $\chi (r)\in 𝕌$, where $𝕌=\{z\in ℂ\mid |z|=1\}$ is a multiplicative subgroup of ${ℂ}^{\ast }$, so $Ĝ\simeq \mathrm{Hom}(G,𝕌)$.

(a)

We know that the set ${({ℂ}^{\ast })}^G$ of the applications from $G$ in ${ℂ}^{\ast }$ is a group for pointwise multiplication. We verify that $Ĝ$ is a subgroup of this group

• $Ĝ\subseteq {({ℂ}^{\ast })}^G$. Moreover If $\chi ,\lambda \in Ĝ$, then for all $g,h\in G$,

  $$\begin{array}{llll}\hfill (\mathit{𝜒𝜆})(\mathit{𝑔h})& =\chi (\mathit{𝑔h})\lambda (\mathit{𝑔h})& \hfill & \\ \hfill & =\chi (g)\chi (h)\lambda (g)\lambda (h)& \hfill & \\ \hfill & =\chi (g)\lambda (g)\chi (h)\lambda (h)& \hfill & \\ \hfill & =(\mathit{𝜒𝜆})(g)(\mathit{𝜒𝜆})(h),& \hfill & \end{array}$$

  so $\mathit{𝜒𝜆}\in Ĝ$.

• Let $\chi$ be any element of $Ĝ$. Then the inverse of $\chi$ in the group ${({ℂ}^{\ast })}^G$ is ${\chi }^{-1}$, defined by ${\chi }^{-1}(g)=1∕\chi (g)(=\overline{\chi (g)})$ for all $g\in G$. Then for all $g,h\in G$,

  $$\begin{array}{llll}\hfill {\chi }^{-1}(\mathit{𝑔h})& =\frac{1}{\chi (\mathit{𝑔h})}& \hfill & \\ \hfill & =\frac{1}{\chi (g)\chi (h)}& \hfill & \\ \hfill & =\frac{1}{\chi (g)}\frac{1}{\chi (h)}& \hfill & \\ \hfill & ={\chi }^{-1}(g){\chi }^{-1}(h),& \hfill & \end{array}$$

  so ${\chi }^{-1}\in Ĝ$.

This shows that $Ĝ$ is a subgroup of ${({ℂ}^{\ast })}^G$.

(b)

Let $G$ be an abelian group. We can write $$\begin{array}{lll}\hfill G\simeq ⟨x_1⟩\times \cdots \times ⟨x_r⟩,\text{where }|x_i|=n_i.& & \hfill \end{array}$$

By Exercise 13, $G$ has presentation

$$\begin{array}{lll}\hfill G\simeq ⟨x_1,\dots ,x_r\mid x_1^{n_1}=\cdots =x_r^{n_r}=1,x_i{x}_j=x_j{x}_i(1\le i\le n,1\le j\le n)⟩.& & \hfill \text{(1)}\end{array}$$

Fix some $i\in [[1,r]]$. Since ${\left(e^{2\mathit{𝜋𝑖}∕n_i}\right)}^{n_i}=1$, and $1^{n^j}=1$ for $j\ne i$, there exists by Exercise 13 a unique homomorphism ${\chi }_i\in Ĝ$ such that

$${\chi }_i(x_i)=e^{2\mathit{𝜋𝑖}∕n_i},{\chi }_i(x_j)=1\text{ if }j\ne i.$$

Then ${\chi }_i^{n_i}(x_i)={\left(e^{2\mathit{𝜋𝑖}∕n_i}\right)}^{n_i}=e^{2\mathit{𝜋𝑖}}=1$, and ${\chi }_i^{n_i}(x_j)=1$ if $j\ne 1$. Since $G=⟨x_1⟩\times \cdots \times ⟨x_r⟩$, we obtain

$${\chi }_i^{n_i}=1.$$

Conversely, if ${\chi }_i^k=1$ for some integer $k$, then ${\chi }_i^k(g)=1$ for all $g\in G$. In particular, ${[{\chi }_i(x_i)]}^k=1$, thus $e^{2\mathit{𝑘𝜋𝑖}∕n_i}=1$, so $e^{2\mathit{𝑘𝜋𝑖}∕n_i}=1$, which gives $2\mathit{𝑘𝜋}∕n_i\equiv 0(\mathit{𝑚𝑜𝑑}2\pi )$ and $n_i\mid k$.

So $n_i$ is the least positive integer among the integers $k$ such that ${\chi }_i^k=1$. This shows

$$|{\chi }_i|=n_i.$$

Since ${\chi }_i^{n_i}=1$ (where $Ĝ$ is abelian), there exists by (1) a unique homomorphism $\phi :G\to Ĝ$ such that $\phi (x_i)={\chi }_i$ for all $i\in [[1,r]]$.

Furthermore, let $\chi$ be any element of $Ĝ$.

Since $x_i^{n_i}=1$, then $\chi {(x_i)}^{n_i}=1$, thus there is some integer $k_i$ such that

$$\chi (x_i)=e^{k_{i}2\mathit{𝑖𝜋}∕n_i}(1\le i\le r).$$

If $g$ is any element of $G$, then $g=x_1^{a_1}\cdots x_r^{a_r}$, thus

$$\begin{array}{llll}\hfill \chi (g)& =\chi {(x_1)}^{a_1}\cdots \chi {(x_r)}^{a_r}& \hfill & \\ \hfill & =e^{a_1{k}_{1}2\mathit{𝑖𝜋}∕n_i}\cdots e^{a_r{k}_{r}2\mathit{𝑖𝜋}∕n_i}& \hfill & \\ \hfill & ={\chi }_1{(g)}^{k_1}\cdots {\chi }_r{(g)}^{a_r}& \hfill & \\ \hfill & =({\chi }_1^{k_1}\cdots {\chi }_r^{k_r})(g),& \hfill & \end{array}$$

so $\chi ={\chi }_1^{k_1}\cdots {\chi }_r^{k_r}$. This shows

$$Ĝ=⟨{\chi }_1,\dots ,{\chi }_r⟩,$$

and consequently $\phi$ is surjective.

Now we prove that $\phi$ is injective. Let $g=x_1^{a_1}\cdots x_r^{a_r}\in \ker <!--\; FUNCTION\; APPLICATION\; -->(\phi )$. Then

$$\begin{array}{llll}\hfill 1& =\phi (g)& \hfill & \\ \hfill & =\phi {(x_1)}^{a_1}\cdots \phi {(x_r)}^{a_r}& \hfill & \\ \hfill & ={\chi }_1^{a_1}\cdots {\chi }_r^{a_r}.& \hfill & \end{array}$$

Then $1=({\chi }_1^{a_1}\cdots {\chi }_r^{a_r})(h)$ for all $\in G$. In particular, for $h=x_i$, we obtain for all $i$

$$\begin{array}{llll}\hfill 1& =({\chi }_1^{a_1}\cdots {\chi }_r^{a_r})(x_i)& \hfill & \\ \hfill & ={\chi }_i{(x_i)}^{a_i}& \hfill & \\ \hfill & =e^{2\mathit{𝜋𝑖}{a}_i∕n_i}.& \hfill & \end{array}$$

Therefore $n_i\mid a_i$, thus $x_i^{a_i}=1$ for all $i$, so $g=1$. This shows that $\phi$ is injective. Hence $\phi$ is an isomorphism, and

$$Ĝ\simeq G.$$