Modern Algebra - Finite Subgroups of Q*

[ESLASL1]

Homework Statement

Find all the subgroups of Q* (set of all non-zero rational #s) under multiplication. Explain how you know that Q* has no other finite subgroups.

Homework Equations

The subgroups must satisfy the properties of association, closure, inverse, and identity.

The Attempt at a Solution

I have determined that Q+ under multiplication is a subgroup (and is not cyclic) but is this the only one? Would I also include multiples? For example, 6^n where n is an integer under multiplication. How do I know there are no other finite subgroups?

 

[Dick]

Neither of the subgroups you've found so far is a finite subgroup. I assume you want to find finite subgroups?

 

[ESLASL1]

You are right. I do want finite subgroups of Q* and neither of the two I mentioned are as such. Thank you for pointing this out. Of course, now I am even more frustrated and unclear as to how to pursue the solution.

 

[Dick]

You are right. I do want finite subgroups of Q* and neither of the two I mentioned are as such. Thank you for pointing this out. Of course, now I am even more frustrated and unclear as to how to pursue the solution.

Is {-1,1} a finite subgroup?

 

[jbunniii]

Hint: If a group contains an element $x$, then it must contain all powers of $x$.

 

[JonF]

For example, 6^n where n is an integer under multiplication

You forgot about inverses. You may want to consider this question in terms of generating sets, it will probably be easier.

 

[Dick]

You forgot about inverses.

The inverse of 6^(n) is 6^(-n). I think if n is an integer, -n is also an integer.

 

[JonF]

I read natural for some reason, you are right.