3
$\begingroup$

I've shown that the two groups $\mathbb{Z_2}\times\mathbb{Z_{12}}$ and $\mathbb{Z_4}\times\mathbb{Z_6}$ are isomorphic, but the map that I constructed is rather messy. Is there a more straightforward way to see this?

I thought perhaps I could say that since $gcd(3,4)=1$, we have $\mathbb{Z_3}\times\mathbb{Z_4}\cong\mathbb{Z_{12}}$, and since $gcd(2,3)=1$, we have $\mathbb{Z_2}\times\mathbb{Z_3}\cong\mathbb{Z_6}$, so our original groups are isomorphic to $\mathbb{Z_2}\times\mathbb{Z_3}\times\mathbb{Z_4}$ and $\mathbb{Z_4}\times\mathbb{Z_2}\times\mathbb{Z_3}$ respectively. Both groups have order 24, and the fundamental theorem of finitely generated abelian groups says that since these group differ only in the order of their factors, they are isomorphic. Hence, we have $\mathbb{Z_2}\times\mathbb{Z_{12}}\cong\mathbb{Z_2}\times\mathbb{Z_3}\times\mathbb{Z_4}\cong\mathbb{Z_4}\times\mathbb{Z_2}\times\mathbb{Z_3}\cong\mathbb{Z_4}\times\mathbb{Z_6}$. Is this argument correct?

$\endgroup$
2
$\begingroup$

Yes, that's correct. You don't have to appeal to the fundamental theorem of finitely generated abelian groups though: $\mathbb{Z}_2\times \mathbb{Z}_3 \times \mathbb{Z}_4$ and $\mathbb{Z}_4\times\mathbb{Z}_2\times\mathbb{Z}_3$ are obviously isomorphic by the map $(a,b,c)\mapsto (c,a,b)$. In general, $G\times H\cong H\times G$. The real work here (and the "messiness" of the explicit map you say you've found) is hidden in the two applications of the Chinese remainder theorem.

$\endgroup$

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.