- #1
jbarrera
- 3
- 0
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.
----------------------------------------------------------------------------------------
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!
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.
----------------------------------------------------------------------------------------
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!