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

**killercatfish** · March 13th 2010, 04:57 PM

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} ?

Thanks in advance.

**Black** · March 13th 2010, 06:53 PM

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

**tonio** · March 13th 2010, 07:00 PM

Originally Posted by Black

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

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...

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

**Drexel28** · March 13th 2010, 10:04 PM

Fundamental theorem of finitely generated abelian groups?

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

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