TT's Maths Thread

[TrueTears]

[IMG]http://img714.imageshack.us/img714/5115/proofgroups.jpg[/img]

Just not too convinced on this proof...

For Case I, it is proving the isomorphism for if the order of G is infinite, however why does the order of G being infinite have anything to do with it specifying that , ? Can't also be a possibility? In other words, there could be 2 exponents which when a is raised to yield the same element in G. So shouldn't Case I have 2 parts, 1. Showing that G is isomorphic to if all the elements in G can be represent with distinct exponents of a. and 2. Show that it's isomorphic if some of the elements in G can be represented with 2 or more exponents of a.

So why has Case I left out a part?


Wait... or is Case I stating that if G is a cyclic group with generator a and the order of G is infinite, then ?

And Case II is stating that if G is a cyclic group with generator a but the order is now FINITE, then there must exist some positive integer m such that . Actually this was just proven in a question a few posts ago lol.

So how do you show that if G is a cyclic group with generator a and the order of G is infinite, then ?

I am not too sure on their proof: "Suppose that ...." because can't this method also be applied and show that even if the order of G is finite then ? Which would then be a contradiction to what I proved a few posts earlier?

For example: Consider WLOG assume h>k, suppose that

So the conclusion is , but we clearly know this is a false statement...

[schnappy]

I can see the real world applications of pure maths presenting themselves in this thread.

[/0]

Hmmm well if generates G and there exists m such that , then just intuitively, can only contain elements, so it can't be infinite cyclic.



Each column represents one element.

[kamil9876]

^ of course you mean btw

this is where it fails for a finite cyclic group. Obvious example is if we consider the group with two element in which

[TrueTears]

ok thanks guys

so that means

1. if G is a cyclic group with generator a and the order of G is infinite, then ?

and

2. if G is a cyclic group with generator a and the order of G is finite, then for some ?

But why is m only restricted to ? I would have thought for 1. it would be for since a is a generator... and same with 2. m should be in as well since a is a generator.


ohhh wait i see, i think the restriction on m is enough if m is in since if G is of finite order and a is a generator then the group G is "symmetric", what i mean is:

since





.
.
.



?

but using the above logic doesnt make sense, consider the cyclic finite group

1 is a generator.

clearly

since

so... im confused lol

[kamil9876]

so as to your question "why is the restriction to ?" well yeah you are right we can instead just restrict it to but so what? we didn't need that to go through the proof.

And I have no idea what you mean by . It doesn't need to hold for a generator . I think what you might have wanted to say is that if is a generator then so is , that is true.

[TrueTears]

ahh yes i finally get it, thanks !

[TrueTears]

[IMG]http://img21.imageshack.us/img21/5066/614ig.jpg[/img]

Just a little unsure on 2 parts in this proof.

First how does imply n divides ms?

Also in the final paragraph, why does <d> contain all the positive m less than n such that (m,n) = d?

Cheers

[kamil9876]

More generally, if then . To see this write with . Notice that this means which means that since .

Of course the more intuitive way of seeing this (and it's good to understand it like this once you get used to it) is look at /0's table (with the MINIMAL positive such that ) Notice that the elements in the SAME column are the same, hence the ones equal to are those in the rightmost columns, Viola.

[TrueTears]

cheers kamil i get it

Also just another question:

Cayley's theorem states that every group, G, is isomorphic to a subgroup of the group of all permutations of G.

I know how to prove this etc however I am just wondering, how do we actually FIND this subgroup? The theorem only tells us the existence of a subgroup but does not tell us how to find it. Is there any slick ways to do so?

Cheers

[kamil9876]

There is a way of finding it if you know the Cayley table. For example look at the table here http://en.wikipedia.org/wiki/Klein_four-group

Notice that each row is a permutation of the elements of the group. Let us encode e,a,b,ab as 1,2,3,4 respectively. Using this encoding the first row reads as:

1 2 3 4

the second is:

2 1 4 3

the third is:

3 4 1 2

the fourth is:

4 3 2 1

So that the subgroup of that is isomorphic to the group consists of the permutations:

1 2 3 4
1 2 3 4

1 2 3 4
2 1 4 3

1 2 3 4
3 4 1 2

1 2 3 4
4 3 2 1

In fact you can use this very idea to write out a proof of Cayley's Theorem.

[TrueTears]

ahhh i see, yeah actually my book went into details of how to find it in latter chapters haha guess i was too eager to find out...


just another question ive thought up while reading about direct products, would say, changing the order of the factors in ANY direct product yield a group isomorphic to the original one?

or more formally, let be groups, then the direct product of the groups, is also a group. Call this group .

Now if we reshuffle the order of the factors and say we have , clearly this would be also be a group, call this group .

Then does ?

[kamil9876]

A general hint for finding isomorphisms: try the obvious.

[TrueTears]

yeah i know how to prove isomorphisms... defining the function and etc but i dono how to define such a function in this case

[kamil9876]

I know so try to define that function, think up of a definition. You get g_2 from g_1 by permuting the factors, so what's an obvious way to get an element from g_1 and changing it to an element of g_2?