3
$\begingroup$

If $G = \mathbb{Z}_{p_1}^{\alpha_1} \oplus ... \oplus \mathbb{Z}_{p_t}^{\alpha_t}$ is it true that every subgroup of $G$ looks like $H_1 \oplus ... \oplus H_t$, where $H_i \leqslant \mathbb{Z}_{p_i}^{\alpha_i}$?

UPD:

  1. $p_i$ are primes (they may be equal).
  2. "looks like" means that every subgroup is isomorphic to $H_1 \oplus ... \oplus H_t$ (I want to know if I can find all subgroups using combintaions of subgroups in $\mathbb{Z}_{p_i}^{\alpha_i}$).
$\endgroup$
  • 2
    $\begingroup$ You might try the minimal case where $G$ is a direct sum of two such cyclic groups. Of course a subgroup of a finite group is finite and a subgroup of an abelian group is abelian, so one certainly expects the Fundamental Thm. of Finite Abelian Groups to be decisive here. $\endgroup$ – hardmath Dec 4 '20 at 18:46
  • 1
    $\begingroup$ Two questions: (1) What does "looks like" mean? Do you only want $H$ to be abstractly isomorphic to $H_1 \oplus \cdots \oplus H_t$, or do you want it to equal $\{(h_1, \dots, h_t): h_i \in H_i\}$ as a subgroup of $G$? (2) What are you assuming about the $p_i$? In particular, do they have to be relatively prime? $\endgroup$ – Ravi Fernando Dec 4 '20 at 18:48
4
$\begingroup$

There is actually a slightly subtle point lurking in your question. Do you want each subgroup of $G$ to be isomorphic to a direct sum of some $H_i$? Or do you want it to be equal to the direct sum of some $H_i$?

In the first case, as Shaun says, you can use the fundamental theorem of finite abelian groups to obtain an affirmative answer; I'll let you work out the details.

In the second case, the answer is negative. For instance, consider the diagonal subgroup $\Delta=\{(0,0),(1,1)\}$ of the finite abelian group $\mathbb{Z}/2\oplus\mathbb{Z}/2$. If $\Delta$ were equal to a direct sum of some $H_1,H_2\leqslant\mathbb{Z}/2$, note that each $H_i$ would have to contain both $0$ and $1$ and hence be equal to all of $\mathbb{Z}/2$. However, $\Delta\neq\mathbb{Z}/2\oplus\mathbb{Z}/2$.

Indeed, $\Delta\cong\mathbb{Z}/2$, so for $\Delta$ to be isomorphic to a direct sum of two subgroups of $\mathbb{Z}/2$ we would need one of the two to be the trivial subgroup.

So, if your question is whether you can "find all subgroups using combinations of subgroups in $\mathbb{Z}_{p_i}^{\alpha_i}$", the answer is true if you add the qualifier "up to isomorphism", but false as written.

$\endgroup$
1
$\begingroup$

Hint: Every subgroup of a finite abelian group is itself a finite abelian group.

Use the Fundamental Theorem of Finite Abelian Groups.

$\endgroup$

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.