1
$\begingroup$

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!

$\endgroup$
  • 1
    $\begingroup$ You're right.$ $ $\endgroup$ – Kenny Lau Jan 21 '18 at 19:02
  • $\begingroup$ the chinese remainder may help here to define a bijection with $\Bbb Z_{12}$ $\endgroup$ – Guy Fsone Jan 21 '18 at 19:15
2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.