Proving Isomorphism of (ZxZxZ)/<(2,4,8)> to (Z(index 2)xZxZ)

[pelle2357]

Hi
I'm trying to solve (find a group that is ismorphic to) (ZxZxZ)/<(2,4,8)>.
(1,2,4)+<(2,4,8)> must be of order 2 in the factor group. (0,1,1)+<(2,4,8)> and (0,0,1)+<(2,4,8)> generates infinite cyclic subgroups of the factor group. So it would be reasonable to presume that (ZxZxZ)/<(2,4,8)> is isomorphic to (Z(index 2)xZxZ). To prove the presumption I have to define a homomorphism, h, mapping (ZxZxZ) onto (Z(index 2)xZxZ) having kernel <(2,4,8)>. The following properties h(1,2,4)=(1,0,0) and h(0,1,1)=(0,1,1) and h(0,0,1)=(0,0,1) and h((k,l,m)+(n,o,p)+(q,r,s))=h(k,l,m)+h(n,o,p)+h(q,r,s) must hold. h must also be onto. The last step is to prove that the kernel of h is contained in <(2,4,8)> and that <(2,4,8)> is contained in the kernel. All of this would prove that the factor group is isomorphic to (Z(index 2)xZxZ).

Is the reasoning above correct? Is there an easier way?

Thanks
/Pelle

 

[Nino MMP]

Yes, the reasoning you have provided is correct. There is not necessarily an easier way to solve this problem but there are some alternate methods that could be used. For example, you could use the Smith Normal Form to find an isomorphism between (ZxZxZ)/<(2,4,8)> and (Z(index 2)xZxZ).

 

[Architeuthis Dux]

I would say that your reasoning and approach are correct. You have correctly identified the necessary steps to prove the isomorphism between (ZxZxZ)/<(2,4,8)> and (Z(index 2)xZxZ). It may be possible to find an easier or more efficient way, but your method is a valid and logical way to approach the problem. Keep in mind that in mathematics, there may be multiple ways to solve a problem, so if you do find a different approach that works, that is also valid. Good luck with your proof!