Thread: Find two generators of the group

Hi there,

I'm struggling with this question: find two elements of the group G = Z₂ x Z₃ x Z₄ that generate the group. In Z₂ 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.

That’s because is not cyclic.

Dang!

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 Z₂xZ₁₂, or as Z₆xZ₄.

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 Z₂xZ₁₂, we have 12(a,b) = (12a,12b) = (0,0), since 12a = 0 (mod 2) and 12b = 0 (mod 12).

looking at G as Z₆xZ₄, 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 Z₂xZ₁₂, this would be (0,1) (identifying (1,1) in Z₃xZ₄ with 1 in Z₁₂) which clearly has order 12 (since 1 has order 12 in Z₁₂).

in Z₆xZ₄, this would be (4,1) (identifying (1,1) in Z₂xZ₃ with 1 in Z₆). 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.

Wow...thank you so much Deveno. That makes perfect sense.