All subgroups of a finite abelian group

Question

It is well known that a finite abelian group $G$ (written additively) can be written as $$ G \simeq \mathbf{Z}_{p_1^{\alpha_1}} \oplus \cdots \oplus \mathbf{Z}_{p_k^{\alpha_k}}, $$ for some not necessarily distinct primes $p_1,\ldots,p_k$. Now, given nonnegative integers $\beta_1,\ldots,\beta_k$ such that $\beta_i\le \alpha_i$ for each $i$, then $$ \mathbf{Z}_{p_1^{\beta_1}} \oplus \cdots \oplus \mathbf{Z}_{p_k^{\beta_k}} $$ is (isomophic to) a subgroup of $G$.

Question: Is there a name for the finite abelian groups $G$ such that all and only subgroups of $G$ are the direct sums above?

Edit: I just noticed it is related related to this question and this other one..

They are those where $p_1,\dots,p_s$ are pairwise distinct. I seem to remember basic groups, but I may be wrong. — egreg, Mar 23 '17 at 22:28
@egreg Thanks! I found basic subgroup on Wikipedia, but it does not seem to be the same.. https://en.wikipedia.org/wiki/Basic_subgroup — Paolo Leonetti, Mar 23 '17 at 23:22
@PaoloLeonetti The subgroup ${(x,x): x \in \mathbb{Z}_p}$ is isomorphic to $\mathbb{Z}$, so I'm not sure that's a counterexample. But you want the subgroups to be generated by elements of the form $(0,\ldots, 1, \ldots, 0)$, right? In that case, your counterexample works. But it wasn't clear to me at first that this was what you meant. — Joshua Ruiter, Mar 24 '17 at 00:57
@JoshuaRuiter Right! However, I think you meant $\mathbf{Z}_p \oplus {0}$ — Paolo Leonetti, Mar 24 '17 at 07:45

score 1 · Answer 1 · answered Mar 24 '17 at 17:46

Haha! These are called (finite) cyclic groups. As egreg remarked, it must be the case that no same prime number $p$ occurs more then once, or there would be a subgroup $(\Bbb Z/p\Bbb Z)^2$ that spoils your property. But then the Chinese remainder theorem applies and your group is cyclic. Conversely any cyclic group is easily seen to have this property, again by the Chinese remainder theorem if you like.

All subgroups of a finite abelian group

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