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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