# Define the dual group $\hat G$ as the set of all homs. from $G$ to $\Bbb C^*$, with pointwise multiplication. Show $\hat G$ is an abelian group.

I don't really understand the nature of $$\hat G$$ as described in this question:

For any group $$G$$, define its dual group $$\hat G$$ to be the set of all homomorphisms from $$G$$ into $$\mathbb{C}^*$$, together with the binary operation of pointwise multiplication of functions.

Show that this binary operation makes $$\hat G$$ into an abelian group.

I understand that any element of finite order would have to be mapped into some element of $$\mathbb{C}^*$$ that's of the same order, which means $$x \in \mathbb{C}^*$$ such that $$x^a = 1$$, but I don't know what those elements look like other than the obvious ($$i, -1, -i$$ and $$1$$).

To emphasize my ignorance, to me the set of all homomorphisms from an arbitrary group into an uncountably infinite set seems like a really general set that would be difficult to analyze.

• The elements of finite order in $\mathbb{C}^*$ are the roots of unity, $e^{2\pi i/n}$.
– lhf
Commented Sep 25, 2013 at 1:31
• Note that an element of finite order isn't necessarily mapped to an element of the same order. For example it can be mapped to $1$. In general it's mapped to an element whose order divides the order of the original element. Commented Sep 25, 2013 at 1:47
• What i suggest is to take $G$ as the cyclic (additive) group with six elements, call them $\{0,1,2,3,4,5 \}$ as usual. Now find the homomorphisms from this group to the multiplicative group of the complex numbers. Commented Sep 25, 2013 at 2:00

You only really need to understand one example to understand the case of any finite group: $\widehat{\mathbb{Z}/n\mathbb{Z}}$.

Indeed, it's not hard to show that $\widehat{A\times B}\cong \widehat{A}\times\widehat{B}$ by the obvious maps (this is just saying that Hom is additive in the first entry). From this and the structure theorem we see that $\widehat{A}$ is just a product of groups of the form $\widehat{\mathbb{Z}/n\mathbb{Z}}$.

Now, if $G$ is a non-abelian group we merely note that

$$\text{Hom}_\mathbf{Grp}(G,\mathbb{C}^\times)=\text{Hom}_\mathbf{Grp}(G^\text{ab},\mathbb{C}^\times)$$

where $G^\text{ab}$ is the abelianization of $G$ (i.e. $G/[G,G]$). Thus, we see that

$$\widehat{G}\cong\widehat{G^{\text{ab}}}$$

So, we really can restrict ourselves to looking at $\widehat{\mathbb{Z}/n\mathbb{Z}}$.

Well, what is a homomorphism $\mathbb{Z}/n\mathbb{Z}\to\mathbb{C}^\times$? It's just specifying an element $z$ of $\mathbb{C}^\times$ for which $1$ maps to, and the only condition on $z$ is that $z^n=1$. Thus, we are really looking at mappings $\mathbb{Z}/n\mathbb{Z}\to \mu_n$ (where the latter is the group of $n^{\text{th}}$ roots of unity). But, it's clear that $1$ can map to any element of $\mu_n$, and thus we see that specifying $\mathbb{Z}/n\mathbb{Z}\to\mu_n$ is really just picking some $e^{\frac{2\pi i r}{n}}$ for which $1$ should map to.

In fact, it's not hard to see that from this that the mapping

$$\mathbb{Z}/n\mathbb{Z}\to\widehat{\mathbb{Z}/n\mathbb{Z}}:r\mapsto \left(1\mapsto e^{\frac{2\pi i r}{n}}\right)$$

is actually an isomorphism of groups!

Thus, as groups, $\widehat{\mathbb{Z}/n\mathbb{Z}}$ is nothing but $\mathbb{Z}/n\mathbb{Z}$, and the isomorphism is very explicit. Following previous statements you can then deduce that $\widehat{G}\cong G^\text{ab}$, and the isomorphism is fairly explicit. In fact, you should prove to yourself that if $\widehat{\mathbb{Z}}:=\text{Hom}(\mathbb{Z},\mathbb{T})$, then $\widehat{\mathbb{Z}}\cong\mathbb{T}$(where $\mathbb{T}$ is the circle group), and so you can extend the above calculation to finitely generated groups $G$.

The above is actually just the very beginning of a rich and beautiful subject. Namely, we can extend the above notion of a dual group from a finite group, or a finitely generated group, to any locally compact abelian group. The correct notion of dual there is the Pontryagin Dual of a locally compact group $G$, which is defined to be the group of continuous homomorphisms $G\to\mathbb{T}$ with the obvious group structure, and with the topology of uniform convergence on compact subsets. There it is not true that $G\cong\widehat{G}$ (we have seen one example of this). But, there is the celebrated Pontryagin duality theorem which states that $\widehat{\widehat{G}}\cong G$ in a natural way for all $G$.

This level of generality is the most illuminating way to understand many of the common notions in mathematics, such as the Fourier transform. In fact, it is ubiquitous in modern mathematics, from analysis (e.g. the Fourier transform) to number theory (the character group of various local and global fields).

• I've actually turned in the assignment by now, but I still want to understand this problem. I'm not really clear on how we know for sure that the homomorphism is exactly the same as the one of the abelianization...in fact, I'm fairly sure that's what the original problem wanted me to prove. Commented Sep 28, 2013 at 3:50
• @Xindaris What is the universal property of the abelinization $G^\text{ab}$? It's the unique group $\text{G}^{\text{ab}}$ with a homomorphism $\alpha:G\to\text{G}^\text{ab}$ such that whenever $A$ is an abelian group, and $\varphi:G\to A$ is a homomorphism, there exists a unique homomorphism $\psi:G^\text{ab}\to A$ such that $\varphi=\psi\circ\alpha$. Can you see how using this helps? Commented Sep 28, 2013 at 7:10

Fix $$G$$. We check the group axioms and then commutativity.

## Closure:

Let $$\varphi, \psi\in \hat G$$. Let $$g,h\in G$$. Then

\begin{align}\varphi\cdot \psi:G&\to \Bbb C\setminus\{0\},\\ k&\mapsto \varphi(k)\psi(k) \end{align}

is clearly a function and

\begin{align} (\varphi \cdot \psi)(gh)&=\varphi(gh)\psi(gh)\\ &=(\varphi(g)\varphi(h))(\psi(g)\psi(h))\\ &=(\varphi(g)\psi(g))(\varphi(h)\psi(h))\\ &=(\varphi \cdot\psi)(g)(\varphi\cdot\psi)(h), \end{align}

so that $$\varphi\cdot \psi\in\hat G$$.

## Associativity:

This follows from associativity of multiplication. Explicitly, for $$\varphi, \psi,\theta\in\hat G$$ and $$g\in G$$, we have

\begin{align} ((\varphi \cdot \psi)\cdot\theta)(g)&=(\varphi \cdot \psi)(g)\theta(g)\\ &=(\varphi(g)\psi(g))\theta(g)\\ &=\varphi(g)(\psi(g)\theta(g))\\ &=\varphi(g)(\psi\cdot\theta)(g)\\ &=(\varphi\cdot(\psi\cdot\theta))(g), \end{align}

but $$g$$ was arbitrary and so

$$(\varphi \cdot \psi)\cdot\theta=\varphi\cdot(\psi\cdot\theta).$$

## Identity:

Consider

\begin{align} \mathbf{1}:G&\to\Bbb C\setminus \{0\},\\ g&\mapsto 1. \end{align}

It is clearly a function. Let $$g,h\in G$$. Then

\begin{align} \mathbf{1}(gh)&=1\\ &=1\times 1\\ &=\mathbf{1}(g)\mathbf{1}(h), \end{align}

so $$\mathbf{1}\in\hat G.$$

Also, we have for any $$\varphi \in\hat G$$ and any $$g\in G$$, that

\begin{align} (\varphi \cdot\mathbf{1})(g)&=\varphi (g)\mathbf{1}(g) \\ &=\varphi(g)1\\ &=\varphi(g), \end{align}

meaning $$\varphi \cdot\mathbf{1}=\varphi$$; similarly, $$\mathbf{1}\cdot\varphi=\varphi$$. Hence $$\mathbf{1}$$ is the identity.

## Inverses:

Let $$\varphi\in\hat G$$. Consider

\begin{align} \varphi':G&\to \Bbb C\setminus\{0\},\\ g&\mapsto 1/\varphi(g). \end{align}

Since $$\varphi(g)\neq 0$$ by definition, $$\varphi'$$ is clearly a function.

Now let $$g,h\in G$$. Then

\begin{align} \varphi'(gh)&=1/\varphi(gh)\\ &=\frac{1}{\varphi(g)\varphi(h)}\\ &=\frac{1}{\varphi (g)}\frac{1}{\varphi(h)}\\ &=\varphi'(g)\varphi'(h), \end{align}

which means $$\varphi'\in\hat G$$.

Furthermore, for $$g\in G$$, we have

\begin{align} (\varphi'\cdot\varphi)(g)&=\varphi'(g)\varphi(g)\\ &=\frac{1}{\varphi (g)}\varphi (g)\\ &=1\\ &=\mathbf{1}(g), \end{align}

which gives $$\varphi'\cdot\varphi=\mathbf 1$$. Similarly, $$\varphi\cdot\varphi'=\mathbf 1$$. Thus

$$\varphi'=\varphi^{-1}$$

is the inverse of $$\varphi$$ with respect to $$\cdot$$.

Therefore, $$(\hat G, \cdot)$$ is a group.

## Commutativity:

Let $$\varphi, \psi\in \hat G$$. Let $$g\in G$$. Then

\begin{align} (\varphi \cdot\psi)(g)&=\varphi(g)\psi(g)\\ &=\psi(g)\varphi(g)\\ &=(\psi\cdot\varphi)(g). \end{align}

But $$g$$ was arbitrary and so

$$\varphi \cdot\psi=\psi\cdot\varphi.$$

Therefore, $$(\hat G, \cdot)$$ is an abelian group. $$\square$$

• 9 years later... a bit late ! ;-)
– user1107523
Commented Jan 16, 2023 at 14:33
• There's nothing wrong with answering an old questions, @AlbertSchrödinberg. I just closed a duplicate (that has now been deleted), linking to this question. I was going to post this answer there but then I found this. My answer filled a gap and so, naturally, I typed it up :) Commented Jan 16, 2023 at 14:44
• This wan't a reproach, don't worry I'm familiar with late answers, it's just funny to imagine the expression of the question asker seeing the answer a decade later
– user1107523
Commented Jan 16, 2023 at 14:59