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:

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*$p_i$ are primes (they may be equal).

*"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 $\mathbb{Z}_{p_i}^{\alpha_i}$).

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

 Use the Fundamental Theorem of Finite Abelian Groups.

