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I want to show that the automorphism group of $C_p^{k}$ is cyclic for an odd prime $p$.

I know that the order of $Aut(C_n)$ is $\phi(n)$ and so the order of $C_{p^{k}}$ is $\phi(p^{k}) = p^{k-1}(p-1)$. If this is prime, then I can conclude it is cyclic. Not sure if this is is the correct route though. Any help or hints much appreciated.

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  • $\begingroup$ The usual approach is to find an automorphism of order $p^{k-1}$, and an automorphism of order $p-1$. Then show they commute. $\endgroup$ – Steve D May 14 '16 at 18:16
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An automorphism of $C_n$ is uniquely determined by the image of $1$ (generator), and this has to be a generator, hence an invertible element. Hence we obtain a bijection $\mathrm{Aut}(C_n)\to (\mathbb{Z}/n\mathbb{Z})^{\times}$. Since we know that the multiplicative group of the integers modulo $n$ is cyclic for $n=p^k$ with $p>2$ prime (and other cases, see here), we have

$$ Aut(C_{p^k})\cong(\mathbb{Z}/p^k\mathbb{Z})^\times \cong \mathrm{C}_{p^{k-1}(p-1)} \cong \mathrm{C}_{\varphi(p^k)} .$$ In general we know when the automorphism group can be cyclic, see When is the automorphism group $\text{Aut }G$ cyclic?

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    $\begingroup$ I think the meat of the question is showing the multiplicative group is cyclic. $\endgroup$ – Steve D May 15 '16 at 1:44

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