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 $\Omega_1(P), \Omega_2(P), \cdots , \Omega_r(P)(=P)$?
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$\frac{G_{1} \times G_{2}}{H_{1} \times H_{2}} \cong \frac{G_{1}}{H_{1}} \times \frac{G_{2}}{H_{2}}$ What does this tell you?? – Riju Jun 10 '17 at 07:15
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Yes, thats right – Rima Jun 10 '17 at 07:23
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then your quotient should be $\mathbb{Z}_{p^{n-1}}$. Isn't it? – Riju Jun 10 '17 at 07:24
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Yes, It was very simple, I edited my question – Rima Jun 10 '17 at 07:27
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"Clearly"...if one knows what $;\Omega_1;$ means. Why don't you define it so that everybody can read your question? – DonAntonio Jun 10 '17 at 08:01
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I defined it now – Rima Jun 10 '17 at 08:03
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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 $2 \leq k \leq n+1.$
Geoff Robinson
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