That’s because is not cyclic.
Hi there,
I'm struggling with this question: find two elements of the group G = Z_{2 }x Z_{3 }x Z_{4 }that generate the group. In Z_{2 }x Z_{3, }which are coprime, (1,1) is a generator with order 6, but this doesn't extend then to (1,1,1) as a generator of G, because it has order 12, not 24 as required, so I can't even find one generator! Can anyone offer any advice? I know how to solve this problem if the numbers were coprime but that's not the case here.
Thanks for any help.
in other words, we actually need to find TWO generators, to be able to describe G = <a,b>. my recommendation is a = (1,0,0) and b = (0,1,1) but a = (1,1,0) and b = (0,0,1) works equally well.
that is, you can regard G as Z_{2}xZ_{12}, or as Z_{6}xZ_{4}.
either way, 12 is the maximum possible order:
12(a,b,c) = (12a,12b,12c) = (0,0,0) (since 12a = 0 (mod 2), 12b = 0 (mod 3) and 12c = 0 (mod 4).
looking at G as Z_{2}xZ_{12}, we have 12(a,b) = (12a,12b) = (0,0), since 12a = 0 (mod 2) and 12b = 0 (mod 12).
looking at G as Z_{6}xZ_{4}, we have 12(a,b) = (12a,12b) = (0,0), since 12a = 0 (mod 6) and 12b = 0 (mod 4).
it turns out we do actually have elements of order 12: (0,1,1) is one such element, since lcm(3,4) = 12.
in Z_{2}xZ_{12}, this would be (0,1) (identifying (1,1) in Z_{3}xZ_{4} with 1 in Z_{12}) which clearly has order 12 (since 1 has order 12 in Z_{12}).
in Z_{6}xZ_{4}, this would be (4,1) (identifying (1,1) in Z_{2}xZ_{3} with 1 in Z_{6}). it isn't so "obvious" that this has order 12, but note that the first coordinate is only 0 every 3rd multiple, and the second coordinate is only 0 every 4th multiple.