Let $G$ be a finite Abelian group that has exactly one subgroup for each divisor of $|G|$. How does this imply that $G$ is cyclic? [duplicate]

Suppose $G$ is a finite Abelian group that has exactly one subgroup for each divisor of $|G|$.

How does this imply that $G$ is cyclic?

By the Fundamental Theorem of Abelian Groups, $G \cong \mathbb{Z}_{p_1 ^{n_1}} \oplus \mathbb{Z}_{p_2 ^{n_2}} \oplus \cdots \oplus \mathbb{Z}_{p_k ^{n_k}}$ where $p_1 , p_2, \cdots , p_k$ are primes (not necessarily distinct).

I know that if $p_1, \cdots p_k$ are distinct, then $ \mathbb{Z}_{p_1 ^{n_1}} \oplus \mathbb{Z}_{p_2 ^{n_2}} \oplus \cdots \oplus \mathbb{Z}_{p_k ^{n_k}}$ is cyclic.

How do I go about showing that $p_1, \cdots, p_k$ are distinct?

Solution 1:

This is probably written somewhere else, but well, you can prove that if there is at most one subgroup for each divisor then the group is cyclic.

If $a \in G$ has order $d$, $\langle a\rangle$ has order $d$ and there are exaclty $\varphi (d)$ generators for it. Let $S_d$ be the set of elements that generate a subgroup of order $d$, which is $\varphi (d)$ if such a subgroup exists (because there is at most one), or $0$ if it does not. Since all elements generate some subgroup,

$$n = |G| = \sum_{d \mid n} S_d \leq \sum_{d \mid n} \varphi (d) =n$$

so $S_d = \varphi (d)$ for all $d$ and in particular $S_n$ is nonempty so $G$ is cyclic.

Note two things: I am using the formula $\sum_{d \mid n} \varphi (d) = n$ which is simple to prove, and you don't require $G$ to be abelian