I see what you're doing. You need to be quite careful here: is not (usually) constructed a subset of , or even of , and it's not a subgroup of either under any conventional operations. It is normally either defined abstractly or defined concretely a quotient of by the subgroup .(Original post by jack.hadamard)
What I thought is, if I labelled subset as , and defined
then I think, I get a cyclic group. I assume the operation is associative, but it follows from the associativity of multiplication modulo.
Then, if I define , I get that
and this is clearly bijective, so it is an isomorphism to .
In this case I get a cyclic group, for all sets with a prime number of elements.
However, I need this to be true for every number of elements, and if it helps, they will be naturals.
Now, is not a group under multiplication modulo , even when is prime (for instance, has no multiplicative inverse modulo ). And it is always a group under addition, even when isn't prime.
However, is a group under multiplication; its elements are with multiplication defined by . But , and the correspondence is not as simple as mapping, say, 1 to 1 and 2 to 2 and 3 to 3 and so on... the isomorphism will mess up the ordering of the integers.
And on this basis it doesn't make a difference that your indices are natural numbers, since if you want them to act multiplicatively then you'll need to screw up their ordering anyway in order to map them into . So you can choose between making them act additively, so for instance , in which case it makes no difference whether or not is prime; or you can choose another group and group operation and choose your isomorphism a different way.
A Summer of Maths
Announcements  Posted on  

TSR looking different to you this week? Find out why here.  02122016 

View Poll Results: Have you studied any Group Theory already?Voters: 74. You may not vote on this poll

 Follow
 541
 23072012 18:42
Last edited by nuodai; 23072012 at 18:45. 
 Follow
 542
 23072012 18:54
(Original post by nuodai)
So you can choose between making them act additively, so for instance , in which case it makes no difference whether or not is prime; or you can choose another group and group operation and choose your isomorphism a different way.
I'll have a look at it now, and check the details to see whether it works. Thanks!
EDIT: I know what you mean by the above though, you didn't waste time to write it all out. I did it quickly, and slightly mispresented it.Last edited by jack.hadamard; 23072012 at 19:01. 
 Follow
 543
 23072012 18:57
(Original post by jack.hadamard)
I see. Well, it was the first one that I picked! So, if I choose them additively, then it seems plausible?
I'll have a look at it now, and check the details to see whether it works. Thanks!
A more critical issue is that not all subsets of are finite, but I'm sure you can work out a way round that... 
 Follow
 544
 23072012 19:06
(Original post by nuodai)
Yes. But you seem to be making it harder than you have to: by labelling the elements of your subset from 1 up to n you have already defined an isomorphism from your subset to , by the method I described two posts ago, since the group structure is inherited from the identification with the elements of .
A more critical issue is that not all subsets of are finite, but I'm sure you can work out a way round that...
It was just an idea, so I have to think through the details, and see how exactly I'm going to apply it.
Well, more specifically, I'm looking at some periodic sequences, so that's not a problem for now.
EDIT:
(Original post by nuodai)
More specifically, if is any subset and is a group with then any bijection induces a group structure on by defining .Last edited by jack.hadamard; 23072012 at 19:18. 
 Follow
 545
 23072012 19:23
(Original post by nuodai)
A more critical issue is that not all subsets of are finite, but I'm sure you can work out a way round that... 
 Follow
 546
 23072012 19:29
(Original post by jack.hadamard)
Found some links: Does every nonempty set admit a group structure on Google. Interesting!Last edited by nuodai; 23072012 at 19:31. 
 Follow
 547
 23072012 19:38
(Original post by nuodai)
.. a famous example being "every set has a wellordering". 
 Follow
 548
 23072012 21:06
Sorry to break up the Group Theory discussion  would like some thoughts on this...
If is continuous everywhere and can I conclude that
My thoughts were along the lines of:
Last edited by Lord of the Flies; 23072012 at 21:19. 
 Follow
 549
 23072012 21:07
One of the books given for Numbers and Sets is Numbers, Sets and Axioms by A. Hamilton.
Nobody suggested books on Set Theory so far? Any comments?
Usually, two sets are defined to be equal if and , or equivalently, if .
Why is this called the Axiom of ``Extensionality", expressed as: ? It does not 'extend' anything. 
 Follow
 550
 23072012 22:12
(Original post by Lord of the Flies)
Does that work?
Spoiler:Show
It sounds like more of a mouthful, but that way you're not skipping over details.Post rating:1 
 Follow
 551
 23072012 22:25
(Original post by jack.hadamard)
Why is this called the Axiom of ``Extensionality", expressed as: ? It does not 'extend' anything.
Background:
A relation on a set can be thought of in a few ways (all are equivalent). One way is as a subset , which means it's a collection of ordered pairs where . (An example of a relation is the set itself, which can be thought of as the set of all 'coordinates'.) Relations generalise functions: a function is a relation for which for each there is exactly one such that , and we denote this by . In this sense, we can think of a relation as a 'multivalued function'. (For instance, the set containing all where or is a relation, but not a function.) But the nittygritty aside, just think of a relation as being a set containing ordered pairs of elements of some set.
Now, if then we say is an 'predecessor' of . So for instance if is a function then you could say (although most people don't) that is an predecessor of for each in the domain of .
The relevant bit:
A relation is called 'extensional' if each element is determined by its predecessors. That is, if for all we have , then . The axiom of extensionality, therefore, is just the assertion that is an extensional relation.Last edited by nuodai; 24072012 at 02:44. Reason: Typo fixed 
 Follow
 552
 23072012 22:30
(Original post by nuodai)
It doesn't mean 'extension' as in 'make something bigger', it means extension as in 'the ability to include more things'. The reasoning goes a bit deeper into relations.
Background:
A relation on a set can be thought of in a few ways (all are equivalent). One way is as a subset , which means it's a collection of ordered pairs where . (An example of a relation is the set itself, which can be thought of as the set of all 'coordinates'.) Relations generalise functions: a function is a relation for which for each there is exactly one such that , and we denote this by . In this sense, we can think of a relation as a 'multivalued function'. (For instance, the set containing all where or is a relation, but not a function.) But the nittygritty aside, just think of a relation as being a set containing ordered pairs of elements of some set.
Now, if then we say is an 'predecessor' of . So for instance if is a function then you could say (although most people don't) that is an predecessor of for each in the domain of .
The relevant bit:
A relation is called 'extensional' if each element is determined by its predecessors. That is, if for all , then . The axiom of extensionality, therefore, is just the assertion that is an extensional relation. 
 Follow
 553
 23072012 22:36
(Original post by kingkongjaffa)
where can I learn what all that notation means? 
 Follow
 554
 23072012 22:44
(Original post by nuodai)
... 
 Follow
 555
 23072012 23:18
(Original post by nuodai)
The relevant bit:
A relation is called 'extensional' if each element is determined by its predecessors. That is, if for all , then . The axiom of extensionality, therefore, is just the assertion that is an extensional relation.
e.g. I don't attribute it any properties other than being a subset of a Cartesian product?
I am still trying to decipher the last bit though.
Is the same as
If for all , we have , then ?
In this case, I get that the axiom of extensionality would be: is and .
Following this logic, would and be extensional relations? 
 Follow
 556
 23072012 23:24
(Original post by jack.hadamard)
I see. Just for clarity, this notion of a ``relation" is as general as it can be.
e.g. I don't attribute it any properties other than being a subset of a Cartesian product?
(Original post by jack.hadamard)
In this case, I get that the axiom of extensionality would be: is and .

 Follow
 557
 23072012 23:51
(Original post by nuodai)
Yes. But, for instance, if we write to mean that is a subgroup of a group (where we consider the groups as being 'up to isomorphism'); then is not extensional (for instance, and have the same subgroups).
Would you recommend Naive Set Theory? Or, perhaps, any other book that might be useful  I notice most of these have no prerequisites.
Comments suggest it is a well structured introduction that covers all axioms of ZF? (It's funny how naive is this book with this axiomatic nature. ) 
 Follow
 558
 24072012 00:00
(Original post by jack.hadamard)
Oh, yes. I wasn't very sure about the meaning of in that case, but took it as being the same object.
Would you recommend Naive Set Theory? Or, perhaps, any other book that might be useful  I notice most of these have no prerequisites.
Comments suggest it is a well structured introduction that covers all axioms of ZF? (It's funny how naive is this book with this axiomatic nature. )
Promise me you won't just do maths this summer? I'm a bit worried. 
 Follow
 559
 24072012 00:15
(Original post by nuodai)
Promise me you won't just do maths this summer? I'm a bit worried. 
 Follow
 560
 24072012 00:38
(Original post by jack.hadamard)
Would you recommend Naive Set Theory? Or, perhaps, any other book that might be useful  I notice most of these have no prerequisites.
Comments suggest it is a well structured introduction that covers all axioms of ZF? (It's funny how naive is this book with this axiomatic nature. )
Write a reply…
Reply
Submit reply
Register
Thanks for posting! You just need to create an account in order to submit the post Already a member? Sign in
Oops, something wasn't right
please check the following:
Sign in
Not got an account? Sign up now
Updated: August 12, 2013
Share this discussion:
Tweet
Related discussions:
 Functional Skills Maths Level 2
 Surina16's journey to successful A Levels!
 Taking A levels during the summer holidays?
 Economics  Course Ambassador Q&A
 Economics  Course Ambassador Q&A
 The "Am I good enough for Investment Banking/Consultancy?" ...
 The "Am I good enough for Investment Banking/Consultancy?" ...
 Underachieving until now but is it too late to pursue a career in ...
 St Andrews Hopefuls 2017
 Alice's road to high grades
TSR Support Team
We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.