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Let $T$ be a finite abelian group. We can consider $T$ the as group $ℤ/nℤ$ or $ℤ/qℤ×ℤ/mℤ$.

My question is: How I can prove tht $T$ is in bijection with a finite set of natural numbers?

That is, I want to prove that there is a bijection between $T$ and a finite set of natural numbers ${1,2,...,q}$ where $q$ is the finite size of $T$.

Safwane
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    What is the group structure on a finite set of natural numbers? – Abel Apr 06 '13 at 10:15
  • The standard summations of natural numbers. – Safwane Apr 06 '13 at 10:17
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    But it would not be closed under addition. – Karl Kroningfeld Apr 06 '13 at 10:17
  • @Chaos Are you sure you mean standard summation? Don't you mean sum modulus some natural number? – Git Gud Apr 06 '13 at 10:18
  • I have changed isomorphism to bijection. Sorry. – Safwane Apr 06 '13 at 10:18
  • @Chaos What's your definition of finite set? – Git Gud Apr 06 '13 at 10:20
  • A set with a finite number of elements. – Safwane Apr 06 '13 at 10:21
  • @Chaos Given the question as it is, I don't understand what that means, to be honest. – Git Gud Apr 06 '13 at 10:22
  • Wikipedia for one defines finite to mean bijection with a finite set of natural numbers... – Abel Apr 06 '13 at 10:23
  • I want to prove that there is a bijection between $T$ and a finite set of natural numbers ${1,2,...,q}$ where $q$ is the finite size of $T$. – Safwane Apr 06 '13 at 10:24
  • @Abel: Yes. This is the definition. – Safwane Apr 06 '13 at 10:26
  • Where $T$ is a finite abelian group or where $T$ is $\mathbb{Z}/n\mathbb{Z}$ say? – Abel Apr 06 '13 at 10:26
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    @Chaos The problem here is the definition of finite set.As Abel pointed out, one possible definition of a set $X$ to be called finite, is that there exists a bijection between $X$ and $[n\textbf{]}$, for some $n\in \Bbb N_0$, where $[n\textbf{]}={0,1,\ldots ,n-1 }$. And if this is your definition, then there's nothing left to prove. – Git Gud Apr 06 '13 at 10:26
  • @GitGud. Yes this is my definition. To Abel: In all cases the groups have finite number of elements. – Safwane Apr 06 '13 at 10:29
  • @Babak S: But I need the expression of that bijection. – Safwane Apr 06 '13 at 10:30
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    @Chaos As I have pointed out, there is nothing left to prove. What you want to prove is the definition of finite set. – Git Gud Apr 06 '13 at 10:30
  • @Chaos That should have been pointed out eariler. Please add that information to the question's body. – Git Gud Apr 06 '13 at 10:32
  • @Chaos `To Abel: In all cases the groups have finite number of elements.' By definition or is this what you want to prove? – Abel Apr 06 '13 at 10:32
  • If $T$ is finite, you can mark its elements by natural numbers as $x_0,x_1,...,x_{|T|}$. – Mikasa Apr 06 '13 at 10:32
  • @Babak S, Abel and Git Gud: Ok I understand the idea. Thanks a lot. – Safwane Apr 06 '13 at 10:35
  • @Abel: Thank you very much for you answer. Then How about the case of $ℤ/qℤ×ℤ/mℤ$ I still confused about it. – Safwane Apr 08 '13 at 08:17
  • By my argument you have a bijection between $\mathbb{Z}/q\mathbb{Z}\times\mathbb{Z}/m\mathbb{Z}$ and ${1,\ldots,q}\times{1,\ldots,m} = {(i,j)|1\leq i\leq q\mbox{ and }1\leq j\leq m}$. It is pretty straight forward to find a bijection from this set to the set ${1,\ldots,q\cdot m}$, define for instance $\psi((i,j)) = (i-1)\cdot m +j$. – Abel Apr 08 '13 at 14:29
  • @Abel: Thank you very much for your valauable comment. – Safwane Apr 08 '13 at 14:30

1 Answers1

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This is too long to be a comment, so I hope it answers your question:

Lets consider $\mathbb{Z}_n = \mathbb{Z}/n\mathbb{Z}$. We want to construct a bijection with the set $\{1,\ldots,n\}$. Any $a\in\mathbb{Z}_n$ is a non-empty subset of $\mathbb{Z}$ that is closed under addition with multiples of $n$. Since it is closed under addition with arbitrarily large numbers, it must contain some positive number. Define $\phi(a)$ to be the smallest positive number contained in $a$.

Suppose $\phi(a)>n$, then $\phi(a)-n$ is positive and smaller than $\phi(a)$, a contradiction. Thus $\phi\colon \mathbb{Z}_n\to\{1,\ldots,n\}$.

Now it remains to show that $\phi$ is a bijection. Clearly $\phi$ is surjective, since $\phi(\{mn+k|m\in\mathbb{Z}\})=k$ for any $k\in\{1,\ldots,n\}$. Furthermore $\phi$ in injective, since $\phi(a)=\phi(b)$ implies that $\phi(a)\in a\cap b$ and the elements of $\mathbb{Z}_n$ partition $\mathbb{Z}$.

Abel
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