# What do subgroups of $\mathbb{Z}_{p_1}^{\alpha_1} \oplus … \oplus \mathbb{Z}_{p_t}^{\alpha_t}$ look like?

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 $$\mathbb{Z}_{p_i}^{\alpha_i}$$).
• 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. – hardmath Dec 4 '20 at 18:46
• 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? – Ravi Fernando Dec 4 '20 at 18:48

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.

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

Use the Fundamental Theorem of Finite Abelian Groups.