# Math Help - Find two generators of the group

1. ## 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 Z3, 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.

2. ## Re: Find two generators of the group

That’s because $G$ is not cyclic.

Dang!

4. ## Re: Find two generators of the group

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.

5. ## Re: Find two generators of the group

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