# 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

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