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I am trying to prove that an abelian $p$-group with only one subgroup of order $p$ is cyclic. Let $G$ be the group of order $p^n$, and let $H$ be the unique subgroup of order $p$, and let $\phi(g)=g^p$ be the $p$-th power homomorphism. I want to show that if $Im(\phi)$ is cyclic, then so is $G$.

$H$ is precisely the kernel of $\phi$, so by the First Isomorphism Theorem, $Im(\phi) \cong \frac{G}{H}$. So there is a generator $gH$ of $\frac{G}{H}$, which thus has order $p^{n-1}$. Then $g^{p^{n-1}} \in H$. If $g^{p^{n-1}} \neq 1$, then $g$ has order $p^n$, making $G$ cyclic. Here's where I get stuck: what prevents all of the generators becoming 1 when raised to the $p^{n-1}$th power?

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  • $\begingroup$ It's not true. ${\mathbb Z}_{p^\infty}$ is a counterexample. $\endgroup$ – Derek Holt Jan 1 '14 at 9:35
  • $\begingroup$ 1. I meant finite groups. I guess I should have said that. 2. What exactly is $\mathbb Z_{p^\infty}$? $\endgroup$ – Nishant Jan 1 '14 at 14:50
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    $\begingroup$ ${\mathbb Z}_{p^\infty}$ is the additive group ${\mathbb Q}_p/{\mathbb Z}$, where ${\mathbb Q}_p$ is the set of rational numbers whose denominator is a power of $p$. It can also be defined as the multiplicative group of all complex $p^k$-th roots of unity with $k \ge0$. $\endgroup$ – Derek Holt Jan 1 '14 at 15:22
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If $g^{p^{n-1}} = 1$, this means that $g^{p^{n-2}}$ is in $H$, contrary to the assumption that $gH$ has order $p^{n-1}$ in $G/H$. (We may assume $n\ge 2$ of course.)

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  • $\begingroup$ Ah, that's where the uniqueness of $H$ comes in. Thanks! $\endgroup$ – Nishant Jan 1 '14 at 3:47

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