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Exercise:Find the generator of $(\mathbb{Z}_4\times\mathbb{Z}_3)$.

I think the generator is $\langle(1_4,1_3)\rangle$ once the $gcd(3,4)=1$ which implies $(\mathbb{Z}_4\times\mathbb{Z}_3)$ is cyclic.

I do not know if I am proceeding correctly as I have never seen a generator of a direct product.

Question:

Is my answer right? If not, then which are the possible generators?

Thanks in advance!

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    $\begingroup$ You're right.$ $ $\endgroup$
    – Kenny Lau
    Jan 21, 2018 at 19:02
  • $\begingroup$ the chinese remainder may help here to define a bijection with $\Bbb Z_{12}$ $\endgroup$
    – Guy Fsone
    Jan 21, 2018 at 19:15

1 Answer 1

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First of all finding the generator of the subgroup is not right, as there isn't a unique generator of it.

Nevertheless as you have already noted the group is cyclic and indeed $\mathbb{Z}_4 \times \mathbb{Z}_3 \cong \mathbb{Z}_{12}$. So now $\mathbb{Z}_{12}$ has $\phi(12) = 4$ generators and those are the positive integers less than $12$, which are coprime to $12$. Those are $1,5,7,11$. These corespond to the elements: $(1,1),(1,2),(3,1),(3,2)$ in our original group. And indeed these are all possible generators of the subgroup.

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