# Prove that an Abelian group of order 6 must be cyclic (without Lagrange's thm)

• Mar 13th 2010, 04:57 PM
killercatfish
Prove that an Abelian group of order 6 must be cyclic (without Lagrange's thm)
I am trying from two different approached. The first is to try and prove that if |a|=2,3,4,5 then it cannot be in the group for some element a in G.

My other approach was going to be to find an arbitrary order 6 group and then work on the proof from there. I dont seem to know what elements would be in there? G={e, a, b, ab, b^2, ab^2} ?

• Mar 13th 2010, 06:53 PM
Black
There are only two groups of order 6 (up to isomorphism): \$\displaystyle D_3\$ and \$\displaystyle \mathbb{Z}_6\$. Since the group in question is abelian, it can't be isomorphic to \$\displaystyle D_3\$.
• Mar 13th 2010, 07:00 PM
tonio
Quote:

Originally Posted by Black
There are only two groups of order 6 (up to isomorphism): \$\displaystyle D_3\$ and \$\displaystyle \mathbb{Z}_6\$. Since the group in question is abelian, it can't be isomorphic to \$\displaystyle D_3\$.

If he can't use Lagrange then I can't see how can he use that there are only two groups, up to isomorhpism, of order 6...(Happy)

I suppose that building a Cayley table, using abelian and a try-and-error process will eventually work, but I can't see a straightfroward way to do it without Lagrange.

Tonio
• Mar 13th 2010, 10:04 PM
Drexel28
Fundamental theorem of finitely generated abelian groups? (Smirk)