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Let $\mathbb{Q}$ denote the group of rational numbers (with addition as the binary operation) and let $\mathbb{Z}$ denote the subgroup of integers. The Pontryagin dual of a group $G$ is the group $G^* = \operatorname{Hom}(G,\mathbb{Q/Z})$. (It is most useful when G is abelian).

  1. Show that $G^*$ is finite if $G$ is a finite group.
  2. Suppose $G = \mathbb{Z}/n$ for some non-negative integer $n$. Show that $G^*$ is isomorphic to $\mathbb{Z}/n$. Compute $G^*$ (up to isomorphism) for the symmetric group on 3 letters.

for 1 I know if $G$ is finite all its elements are of finite order, and $\mathbb{Q/Z}$ has infinite elements, all of finite order. But I don't know where to go from there or if that's the right idea to begin with. First part of 2 sounds like it follows from similar work that would be done in 1, right now though all I can say is that $G^*$ would be finite. For the second part of 2 I know how the $S_3$ group works but I don't know how to go about finding all the homomorphisms of it to $\mathbb{Q/Z}$, except again, 1 tells me it would be a finite amount. Any help would be appreciated.

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+1 for showing your thoughts about the problem, it is something that can be frustratingly rare. –  Zev Chonoles Nov 11 '11 at 6:26
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Do you mean $\mathbb Z / n\mathbb Z$? +1 for showing thoughts too, I love when people do that. –  Patrick Da Silva Nov 11 '11 at 6:40

3 Answers 3

up vote 5 down vote accepted

Every element of $\mathbb{Q}/\mathbb{Z}$ can be written uniquely in the form $q+\mathbb{Z}$ for $q\in [0,1)$ (if this isn't obvious, you should take a moment to prove it). So, let $G=\{0,g_1,g_2,\cdots,g_n\}$ be a finite group, and denote $n_i:=\vert g_i \vert$. Then, for all $f\in G^*$, $n_i f(g_i)=0+\mathbb{Z}$. Write $f(g_i)=\frac{a_i}{b_i} + \mathbb{Z}$ with $a_i,b_i\in \mathbb{N}$, $a_i<b_i$, and $gcd(a_i,b_i)=1$. Then, in $\mathbb{Z}$, we have $b_i \vert a_i n_i$ which implies $b_i \vert n_i$. Since there are only finitely many positive integer divisors of $n_i$ and only finitely many corresponding choices for $a_i$, there are a finite number of ways to choose $f$. That takes care of the first question.

For the second, I'll bet you can show that the map sending 1 to $\frac{1}{n}+\mathbb{Z}$ is a generator for $G^*$ by using an argument similar to the one above.

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Aww, now my hint is ruined. =P Good work –  Patrick Da Silva Nov 11 '11 at 6:43
    
Thanks, so I get 1 now, but I'm still having problems with 2. I think I'm just having a hard time grasping how to define mappings between groups, especially when one of the groups is a group of functions, that's just hard for me to visualize. I can see how the map sending 1 to 1/n + Z would generate G*. If you 'multiplied' that function by itself you'd get the map sending 2 to 2/n + Z and that would keep going until you got to the map sending n to n/n + Z, which is the same as the mapping for 0. Just not sure how to use that to prove the isomorphism. –  Ryan Nov 11 '11 at 7:49
    
@Ryan - If $f,g\in G^*$, then we can add the maps together by declaring $(f+g)(x)=f(x)+g(x)$. This isn't just a cutesy notation whereby we "distribute the $x$": given two maps $f,g\in G^*$, we're defining a new map, called $f+g$. How is $f+g$ defined? Well, for any $x\in G$, we simply add the images of $x$ under $f$ and $g$. You should prove that $f+g$ is a homomorphism. What is the identity element of $G^*$--ie what element $e\in G^*$ is there such that $e+f=f$ for all $f\in G^*$? What about inverses? –  user5137 Nov 11 '11 at 8:01
    
ok, so f+g is a homomorphism because (f+g)(x+y) = f(x+y) + g(x+y) = f(x)+f(y)+g(x)+g(y) (because f and g are homs) = f(x+y)+g(x+y). The identity of G* would be the map that maps everything to Z, and given an f in G* and a g in G, if f(g)= g/n + Z, the inverse, f^-1, would map g to (n-g)/n + Z. I'm just not sure what all that proves for the question. I feel like I'm missing something very obvious. –  Ryan Nov 11 '11 at 9:01
    
Also trying to figure out a way to show all the homomorphisms of S3 to Q/Z, not sure how to besides just guessing at them. Only found one so far. if S3={(), (12), (23), (13), (123), (132)} then (), (123), (132) -> Z and (12), (23), (13) -> 1/2+Z should be a hom, and then there's the trivial one where everything maps to Z. Not sure if there can be others, since (12),(23), and (13) are all order 2, and 1/2+Z is the only order 2 subgroup of Q/Z, so I can't have them map to anything else can i? Not sure if I'm even going about this the right way, as usual. –  Ryan Nov 11 '11 at 9:52

I got a hint for 1. Yes, $\mathbb Q / \mathbb Z$ has infinitely many elements, but how many elements does it have of order $n$, for given $n$? (Hence these would be the possibilities for elements of $G$ to be mapped to.)

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Someone cares to explain the downvote?... –  Patrick Da Silva Nov 11 '11 at 7:04
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Okay now this is just bothering me ; what's up with all the downvotes?? I was the first to give an answer and I was just giving a hint! It's not because there are other answers that this one must be downvoted ... –  Patrick Da Silva Nov 11 '11 at 7:18
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Third downvote with no explanation : we should be able to downvote downvotes for bad-downvoting. –  Patrick Da Silva Nov 11 '11 at 7:29
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Dear Patrick, +1; this is certainly the key point. Regards, –  Matt E Nov 11 '11 at 14:09
    
Weirdest answer ever. Thanks Matt. How can I get 7+ and 3- on the same answer?... this is so weird. –  Patrick Da Silva Nov 11 '11 at 18:52

Here's an alternate solution to Jack's. Suppose first that $G$ is abelian, and decompose it as a direct sum of cyclic groups. Since $\hom(A \oplus B, H) \cong \hom(A, H) \oplus \hom(B, H)$ for any abelian groups $A,B,H$, we are reduced to showing that $2$ is true. Now it suffices to check directly that there are $n$ possible images for the generator of $\mathbb{Z}/n\mathbb{Z}$ (they are $0/n, 1/n, \dots, (n-1)/n$, as Patrick hints to). In fact, this shows that for a finite abelian group $G$, we have $G\simeq G^*$ ('$\simeq$' denotes non-canonical isomorphism).

Now if $G$ is not abelian, notice that by the fact that the abelianization functor is left adjoint to the forgetful functor $\mathbf{AbGrp} \to \mathbf{Grp}$, we have a canonical isomorphism

$$\hom(G_{Ab}, \mathbb{Q}/\mathbb{Z}) \cong \hom(G, \mathbb{Q}/\mathbb{Z})$$

which shows that $\hom(G, \mathbb{Q}/\mathbb{Z})$ is finite for every finite group $G$ (in fact, for every group $G$ whose abelianization is finite).

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I really need to get acquainted with category theory.. I hate it when you pull out stuff like that and I'm craving to understand what you're saying. –  Patrick Da Silva Nov 11 '11 at 7:05
    
Which part don't you understand? I can supply more details if you want. In this context adjointness simply means the following: there is a bijection between the set of morphisms from some group $G$ to some abelian group $H$, and the set of morphisms from $G_{Ab}$ to $H$. In other words, every mapping from $G$ to some abelian group factors uniquely through the quotient from $G$ to $G_{Ab}$. –  Bruno Joyal Nov 11 '11 at 7:13
    
Hm. What you're trying to explain with that forgetful functor is that for a morphism from $ \varphi : G \to H$ with $H$ abelian, we can factor $\varphi = \varphi_1 \circ \varphi_2$ where $\varphi_1 : G \to G/[G,G]$ and $\varphi_2 : G/[G,G] \to H$? –  Patrick Da Silva Nov 11 '11 at 7:24
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Yes! (But that should read $\varphi_2\circ \varphi_1$.) In fact $\varphi_2$ is uniquely determined by $\varphi$, and the map $\varphi \mapsto \varphi_2$ is a bijection between the corresponding hom-sets. –  Bruno Joyal Nov 11 '11 at 7:28
    
Uh, yeah sure. That was just typo. =) –  Patrick Da Silva Nov 11 '11 at 7:30

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