Abstract Algebra Proof: Groups

1. Oct 18, 2011

jbarrera

Abstract Algebra Proof: Groups....

A few classmates and I need help with some proofs. Our test is in a few days, and we can't seem to figure out these proofs.

Problem 1:
Show that if G is a finite group, then every element of G is of finite order.

Problem 2:
Show that Q+ under multiplication is not cyclic.

Problem 3: Prove that if G = <x> and G is infinite, then x and x^-1 are the only generators of G.
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For problem 2, this is what we have so far:

Suppose by contradiction that (Q+, *) is a cyclic group. Then there exist an element x = a/b, where a and be are in Z+ and (a, b) = 1 such that Q = { (a/b)^n | n in Z }. Let us consider a/(2b).

Then,

a/(2b) = (a/b)^n

1/2 = (a/b)^(n-1)

From here we may consider the following cases:

n = 1: 1/ 2 = 1 which is a contradiction.
n > 1: (a/b) = 1 / 2^(1/n-1) which is a contradiction.
n < 1: (a/b) = 1 / 2^(1/n-1) which is a contradiciton.

Therefore, (Q+, *) is not a cyclic group.

Our professor said, we need to show a little bit more work, but we are not sure. He says our proof is not quite clear.

Any help is greatly apprecaited. Thank you!!!

2. Oct 18, 2011

Citan Uzuki

Re: Abstract Algebra Proof: Groups....

Why is this a contradiction? It clearly holds if, for instance, a/b = 1/2 and n=2. So to make this strategy of proof work, you have to find a way to exclude this possibility as well.

3. Oct 18, 2011

Robert1986

Re: Abstract Algebra Proof: Groups....

OK, for the first one, let's think about what it means for an element to not have finite order. This means that if g is in G, then you can multiply g by itself as many times as you want and never get back to the identity in G. Now, in multiplying g by itself like this, one of two things can happen: a)eventually you will get a repetition (i.e. g^k = g^m for some k,m) or b)you will never get a repetition, no matter how many times you multiply g by itself.

Now, clearly (b) can't be true (but be sure to explain why this is so). This leaves (a). Now, given this, do you see how to prove that g has finite order?

4. Oct 18, 2011

Robert1986

Re: Abstract Algebra Proof: Groups....

For the second one, I would suggest using the fact that if p/q is a generator of Q+, where (p,q) = 1, then there are some members of Q who have some prime factors that neither p nor q have.

5. Oct 18, 2011

jbarrera

Re: Abstract Algebra Proof: Groups....

I see how I can prove (a):

I can say,
Let g be an element in G which does not have finite order. Since (g)^k is in G for each value of k = 0, 1, 2, ...., we can conclude that we may find integers m and n such that (g)^m = (g)^n. If m > n, then g^(m-n) = e. Thus, m-n is the order of G and is finite. Similarly, n > m.

Is that correct?

I don't see why part b would never happen thought. How could I explain that?

6. Oct 18, 2011

micromass

Staff Emeritus
Re: Abstract Algebra Proof: Groups....

For the third one. Let y be another generator. Then

$$y^n=x$$

for a certain n. But x is a generator thus

$$x^m=y$$

for an m. Put all this information together...

7. Oct 18, 2011

Robert1986

Re: Abstract Algebra Proof: Groups....

Yes, this is the gist of it. However, you don't know that m-n is the order of g; all you know is that m-n is greater than or equal to the order of g. But, since m-n is clearly finite, then the order of g is clearly finite.

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Well then, it is probably time to take a little break from studying algebra. What if you could multiply g by itself over and over and over again and never have a repeat? Wouldn't this suggest that there is an infinite number of elements in the group?

8. Oct 20, 2011

wisvuze

Re: Abstract Algebra Proof: Groups....

if you are allowed to use Lagrange's theorem, part a) follows easily