Is H a Subgroup if xy Belongs to H for All x, y in H, and Why Isn't Q Cyclic?

[jacobrhcp]

[SOLVED] group theorem problems

Homework Statement

1) suppose H is a finite subset of a group G. Prove that H is a subgroup of G if and only if xy belongs to H whenever x and y belong to H
2) show that Q is not cyclic, (where Q is the group of rationals with addition.)

The Attempt at a Solution

1) if H is a subgroup, then if x,y are in H, so is by hypothesis xy, because H is a subgroup and therefore closed.

Now conversely, if xy is an element of H when x,y are in H, then x*x is in H
Let n be the number of elements in H, then x^n is in H
Now because H is finite, there has to be an element such that x^n+1=x^m, with n+1>m. Otherwise there would be n+1 distinct elements in a set with n elements. I didn't get any further then this.

I suppose if for elements of H a*b=a*c would imply that b=c, then x=x^n+1-m and so e:=x^n-m. Then I can find and inverse too without many problems, and because H is already know to have an associative operation, and H is closed by hypothesis, H is a subgroup. But is this true?

2) Suppose there is an element x such that <x>=Q. Then x/2 is too in Q, and is not in <x>. But I don't find this a very convincing proof by contradiction.

 

[CompuChip]

1) Are you sure you aren't missing an inverse sign somewhere? As in: H is a subgroup of G iff for all x and y in H, $x y^{-1} \in H$ ? There was a case where you could do without the inverses (G abelian?), but in general you need it.

2) Are Q the rationals, or the rationals - 0 or the quaternions or ...? A group consists of a set and an operation, what is the operation (addition, multiplication, ...).

 

[NateTG]

In part 1, you can get a lot of mileage out of the fact that $G$ is a group, so every element of $H$ has some order in $G$.

Compuchip: Since $H$ is finite, closure over multiplication is sufficient.

 

[jacobrhcp]

compu: for 2) H is finite, which makes the difference. And Q are the rationals (with addition).

nate, what exactly do you mean by 'all elements having some order in G', and what is it's use?

 

[CompuChip]

compu: for 2) H is finite, which makes the difference. And Q are the rationals (with addition).

You're right, I seem to be quite rusty in this subject.

nate, what exactly do you mean by 'all elements having some order in G', and what is it's use?

He means that for every element x in G, there is some number n such that x^n = e (not necessarily the same n for all x though).

The only thing you know is that H is non-empty, so there must be at least one element, x. Now use what nate told you. You will be able to fulfill at least two of the group axioms for H in one blow.

 

[NateTG]

...
He means that for every element x in G, there is some number n such that x^n = e (not necessarily the same n for all x though)...

Actually, $G$ can admit elements of infinite order since it is not necessarily finite. (Of course, those elements can't be part of $H$.)

 

[ircdan]

for 1) take any x in H, then x, x^2, x^3, ... are all elements of H, but H is finite, so at least two of these must be the same, say x^j = x^k for some j != k. You should be able to take it from here.

 

[jacobrhcp]

ircdan, the problem with that argument is that you are not sure that xa=xb implies that a=b. But I think I got it now, thanks everyone!

Anyone got any tips to show Q is not cyclic?

 

[ircdan]

2) Suppose there is an element x such that <x>=Q. Then x/2 is too in Q, and is not in <x>. But I don't find this a very convincing proof by contradiction.

It's almost ok! You need to rule out 0 at the beginning and this is obvious since <0> != Q. So suppose Q was cyclic, then there is x !=0 in Q s.t. <x> = Q. Then x/2 is in Q, but x/2 is not in <x>.
This is enough, because <x> = {nx | n in Z} and x/2 is not of the form n*x for n in Z.

Now if you are not convinced, suppose that x/2 was in <x>, then x/2 = nx for some n in Z.

Then, x= (2n)x, so (2n-1)x = 0(ie, this means x + x + ... + x = 0 (2n-1) times), so x has finite order, but only 0 has finite order in Q, so x = 0, a contradiction.

 

[ircdan]

ircdan, the problem with that argument is that you are not sure that xa=xb implies that a=b. But I think I got it now, thanks everyone!

Anyone got any tips to show Q is not cyclic?

There is no problem there. If xa = xb in H, then this is an equation in G, so a = b. In our case everything is a power of x and x is in H, so everything is in H by hypothesis, so our inverse lies in H. It's a good trick to keep in mind!

 

[jacobrhcp]

thanks, and solved!