TT's Maths Thread

[kamil9876]

what happened on Sunday?

Hint: distribute the

[TrueTears]

what happened on Sunday?

Hint: distribute the

i replied to u on msn, lol my friend didnt reply me, didnt end up going on sunday just worked for parents. maybe sometime this week... ill text u

ok ill try distributing it

[TrueTears]

o i got it, thanks kamil

[IMG]http://img684.imageshack.us/img684/7002/expanding.jpg[/img]

How would I do this?

[TrueTears]

[IMG]http://img607.imageshack.us/img607/2628/assignment5.jpg[/img]

Just wondering for the first one, it should be false right? Because for H to be a subgroup of G it has to satisfy 3 conditions.

1. H has to be closed under *
2. The identity element e of G is in H
3. for all a in H, it is true that the inverse of a is in H.

So H only satisfies 1. But I can't seem to find a counter example...

Also for the 2nd part of the question, I also can't seem to find a counterexample again...

Furthermore, just a question that arose when I thought about the 2nd part of the Q, if a group G has finite elements then does each subgroup of G also have finite elements? Or can a group G that has a finite order have a subgroup that has an infinite order?

Cheers.

[kamil9876]

Furthermore, just a question that arose when I thought about the 2nd part of the Q, if a group G has finite elements then does each subgroup of G also have finite elements? Or can a group G that has a finite order have a subgroup that has an infinite order?

What is a subgroup? it is a subset first of all, so of course this is true.

The other quetions are interesting. The first one is actually TRUE

Because look, if H is non-empty then it must contain some element , and since it is close under the operation it must contain: , But for some $n$  (as is finite) Hence it contains the identity. Likewise if then for some and so .

The second I leave to you. In fact I remember solving some other problems like this of the form "finite set with certain conditions => it is a group" and there are stronger results.

[TrueTears]

Just a few questions about your proof.

Firstly, if H is closed under the operation of G, then does that mean for any element , then for all , or is the restriction on , ?

When you tried to show that the identity element of G is also in H, there's a problem.

Since H is a of finite order, then there exists some element such that for some

That means H contains a identity element. However we still haven't shown that this identity element is the same identity element of G?

Since we don't know if there exists a such that in G because we don't know if G is of finite order or not. We know that the subset H is finite but this does not imply G is finite. G can be an infinite set.


Also just to generalise, if we have a group G (does not necessarily have to be cyclic) and consider an element g in G. If for some then G is of finite order. And the converse is also true, ie, if G is of finite order, then there exists some g in G such that for some .

Is that true?

[kamil9876]

Yes sorry I read it slightly wrong(i thought it said is finite), see the new post I changed

(as G is finite)

to

(as H is finite)

and now the proof is correct.

Firstly, if H is closed under the operation of G, then does that mean for any element g \in H, then g^n \in H for all n \in \mathbb{Z}, or is the restriction on n, n \in \mathbb{Z}^+?

Yes correct we can only infer it for strictly positive integers , but that is no issue in my proof.

Basically we know that for any element there are ONLY two options:

(1) are all distinct

(2) for some positive integer

(I think you even asked about this theorem some posts ago)

Now obviously (1) cannot occur as is finite. Hence (2) occurs, that is my proof.

That means H contains a identity element. However we still haven't shown that this identity element is the same identity element of G?

No, you see is just some random subset of and my argument above is looking at as elements of with the operations from . (and then noticing they are also in )

Also just to generalise, if we have a group G (does not necessarily have to be cyclic) and consider an element g in G. If g^m = e for some m \in \mathbb{Z}^+ then G is of finite order. And the converse is also true, ie, if G is of finite order, then there exists some g in G such that g^m = e for some m \in \mathbb{Z}^+.

Is that true?

Actually your converse is true (It is clearly true by the pigeonhole principle, and you can take to be anything you don't just have to say "there exists", it holds "FOR ALL g" ie an even stronger statement)

The original statement is false though. Any easy counterexample is this:

(infinite product). It is an infinite group but every element is of order 2.

[TrueTears]

Ah okay, i get it now, but also just want to make clear of a few things:

If G is a cyclic group and has generator g, then:

1. , if then G is of infinite order. The converse is also true.

2. If for some then G is of finite order. The converse is also true.

Now i rmb talking to you about these 2 facts and we proved both were true. Now I wanna ask, what if we don't know whether G is cyclic or not and we pick a element g from G such that g might not be a generator of G.

Then 1. and 2. are not true anymore right? But we can adjust it to:

1. If G is of infinite order, then

2. If G is of finite order, then for some

correct?


Also for the 2nd part, is this right?

Counterexample: Let the group G be . Let H be the infinite subset which is closed under +. However the identity element 0 of is not in thus H is not a subgroup of G.


Also just a new question, show that a group with only a finite number of subgroups must be finite.

I've made a start on this question but I just don't quite know if it's heading in the right direction lol.

Assume that a group G is of finite order. Let be an element in G.

Consider the subgroup of G. If is of infinite order, then it is isomorphic to and thus has an infinite number of subgroups (Since all subgroups of are of the form for ) and thus G has an infinite number of subgroups, contradicting our assumption that G is of finite order.

So we can assume that is of finite order. If is of finite order then it is isomorphic to

Thus if then G is isomorphic to and since has a finite number of subgroups, G must also has a finite number of subgroups.

Now this is where I am stuck. First, what if is not a generator for G? And also I just "assumed" has a finite number of subgroups, how can you actually prove that has a finite number of subgroups?

[kamil9876]

Then 1. and 2. are not true anymore right? But we can adjust it to:

1. If G is of infinite order, then g^m \neq e \ \forall m \in \mathbb{Z}^+

2. If G is of finite order, then g^m = e for some m \in \mathbb{Z}^+

correct?

2 is true I verified it in previous post. 1 is not true I showed you a counterexample in the previous post.

Also for the 2nd part, is this right?

Counterexample: Let the group G be <\mathbb{Z}, +>. Let H be the infinite subset \mathbb{Z}^+ which is closed under +. However the identity element 0 of \mathbb{Z} is not in \mathbb{Z}^+ thus H is not a subgroup of G.

yeah.


As for the next question. Decent start.

Here is a hint: You are right, what if was not a generator? then there is some not in , so would be a DIFFERENT subgroup. Now what if there was a that was not in and not in etc. etc. (can this argument go on forever?), what happens when it stops?

[TrueTears]

Cheers

but what counter example? You mean

... (infinite product). It is an infinite group but every element is of order 2.

? what do you mean every element is of order 2? you didn't show that an element g in that group can produce for some

Yeah for the other question, i thought about that as well, but look if we picked a g2 not in g1, it would certainly produce another subgroup, however there could also be elements IN <g_1> that could also produce another subgroup. eg consider Z_18, look at <1>, if we picked 2 from <1> and formed <2> we make another subgroup? Or is that already taken into consideration because like i said <g_1> is isomorphic to some Z_n and thus it has a finite number of subgroups. (is that reasoning right?) so if we had <1> for Z_18, then the other subgroups for <1> is already "taken care" of since we know Z_18 has a finite number of subgroups. So the question remains how do you show Z_n has a finite number of subgroups?


Now continuing on with the question, that means if we pick a g2 not in <g1> and form <g2>, how do you know <g2> would be a different subgroup? How do you define 'different'?

like in my previous example for Z_18, if we had <1>, then <2> is also a different subgroup, ie, <1> = {0, 1, 2, 3, ..., 17} and <2> = {0, 2, 4, ..., 16} theyre both different subgroups of Z_18 yet

So if we assume I don't get what happens if we continue this argument, can't see it clearly...

[kamil9876]

? what do you mean every element is of order 2? you didn't show that an element g in that group can produce g^m = e for some m \in \mathbb{Z}^+

Yes every element is of order 2, if we take any element such as then, .

As for the question, it is essential that you know what it means for two subgroups to be different in order to understand the statement "finitely many subgroups". Two subgroups are said to be the SAME if they are the same set.

Anyway you can do it like I suggested before but I realised there is an easier way: As there are only finitely many subgroups of , there are only finitely many cyclic subgroups of . Now as we saw before each cyclic subgroup of must be finite (for if it was infinite then it has infinitely many subgroups), hence G is a union of finitely many finite cyclic subgroups (each element is in some subgroup (possibly more than one), ie the cyclic subgroup generated by itself). Hence G is finite.

[/0]

Suppose has subgroups. If each subgroup is finite and the largest of them has elements (since you proved the subgroups must be finite) then has elements.
Suppose there was an element of which was not part of a subgroup. Then it would not be in its own cyclic group, which is impossible.

[TrueTears]

? what do you mean every element is of order 2? you didn't show that an element g in that group can produce g^m = e for some m \in \mathbb{Z}^+

Yes every element is of order 2, if we take any element such as then, .

As for the question, it is essential that you know what it means for two subgroups to be different in order to understand the statement "finitely many subgroups". Two subgroups are said to be the SAME if they are the same set.

Anyway you can do it like I suggested before but I realised there is an easier way: As there are only finitely many subgroups of , there are only finitely many cyclic subgroups of . Now as we saw before each cyclic subgroup of must be finite (for if it was infinite then it has infinitely many subgroups), hence G is a union of finitely many finite cyclic subgroups (each element is in some subgroup (possibly more than one), ie the cyclic subgroup generated by itself). Hence G is finite.

Thanks, well i am still not quite sure how to complete the method before, can you answer the queries i have about it? They are the following:

When we said that <g1> = G, then we are done, is that because <g1> is isomorphic to  where n is the order of <g1> and since  has a finite number of subgroups, then so does G? But how can we actually prove that  has a finite number of subgroups?

So if we consider a g2 not in <g1> then <g2> will be another DIFFERENT subgroup, how can you show <g2> is a DIFFERENT subgroup, ie <g1> and <g2> are not the same set?
Is it because if we assume <g1> and <g2> are the same set then g2 must be in <g1> contradicting the fact that we assumed g2 is not in <g1>?

We know that both <g1> and <g2> are finite sets (because of their isomorphisms to for some n). However I am still a bit lost on how to complete the proof.

First why don't we also consider elements IN <g1>? Since these elements could also produce a different subgroup, ie, consider the subgroup <2> of . If we consider the element 6 in <2>, then <6> is a different subgroup as well...

Second, I don't see how picking g1, g2, g3... and then forming cyclic groups out of these elements suffices to show G is finite.


Can you please explain the loose ended questions above?

[TrueTears]

there is an easier way: As there are only finitely many subgroups of , there are only finitely many cyclic subgroups of . Now as we saw before each cyclic subgroup of must be finite (for if it was infinite then it has infinitely many subgroups), hence G is a union of finitely many finite cyclic subgroups (each element is in some subgroup (possibly more than one), ie the cyclic subgroup generated by itself). Hence G is finite.

Also for your new method, i have a few questions, how does there being finitely many subgroups of imply that there are finitely many cyclic subgroups of ?

Also why is G a union of the finitely many finite cyclic subgroups? How do you know that each element of G is in some subgroup? Aren't you basing this on the assumption that each element of G is in at least one subgroup and the subgroup have to be of finite order. What if the subgroup is not of finite order? (you are assuming that G has finitely many subgroups, but that doesn't imply each subgroup has a finite order) What if an element of G is not in any of its subgroups?

[kamil9876]

how does there being finitely many subgroups of G imply that there are finitely many cyclic subgroups of G?

The cyclic subgroups are subgroups. A subcollection of a finite collection of objects is finite, that is obvious?

How do you know that each element of G is in some subgroup

Answer:

(each element is in some subgroup (possibly more than one), ie the cyclic subgroup generated by itself)

AKA so each element is in some cyclic subgroup.

What if the subgroup is not of finite order?

If the CYCLIC subgroup is not of finite order then it has infinitely many subgroups k=1,2,3,4.. (as you showed here

If <g_1> is of infinite order, then it is isomorphic to \mathbb{Z} and thus has an infinite number of subgroups