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Rationals

Name: Anonymous 2008-07-06 16:12

Prove that any finite multiplicative group of the rationals is cyclic.

I did it a while ago, and I can't remember how.

Name: Anonymous 2008-07-06 17:13

You can do it for a finite multiplicative group of any field. Use The Structure Theorem for finitely generated Abelian groups, and consider a polynomial of suitable degree.

Name: Anonymous 2008-07-08 23:32

x*x

Name: Anonymous 2008-07-09 12:06

>>2
I forgot to use "subgroup", not group.

>>3
You're right, we should state we're considering commutative algebras.

Name: Anonymous 2008-07-09 17:13

>>4

I thought a field was necessarily commutative....

Wiki says that's the modern usage. Although a lot of things get defined in ways that make them convenient.

Name: Anonymous 2008-07-10 10:12

>>5
Ok, I thought it might be. It's implicit then, and we shouldn't really have to state it.
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