On the subgroup $\Omega_1(P)$ of a $p$-group

Question

Let $P \cong \Bbb{Z}_{p^n} \times \Bbb{Z}_p$ be an abelian $p$-group of order $p^{n+1}$. We konw $\Omega_1(P)= \langle x \in P \mid x^p=1 \rangle$. Clearly, $\Omega_1(P) \cong \Bbb{Z}_{p} \times \Bbb{Z}_p$ and $P/ \Omega_1(P)$ is cyclic. Why all non-cyclic subgroups of $P$ are $\Omega_1(P), \Omega_2(P), \cdots , \Omega_r(P)(=P)$?

Answer 1

A finite Abelian $p$-group $X$ is non-cyclic if and only if $X$ contains a subgroup isomorphic to $\mathbb{Z}_{p} \times \mathbb{Z}_{p}.$ For your group $P,$ the only available such subgroup of order $p^{2}$ is $\Omega_{1}(P).$ Hence the non-cyclic subgroups of $P$ are precisely the subgroups which contain $\Omega_{1}(P).$ Since $P/\Omega_{1}(P)$ is cyclic, it has a unique subgroup for each divisor of its order, so there is just one non-cyclic subgroup of $P$ of order $p^{k}$ for $2 \leq k \leq n+1.$