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I am a graduate student of Mathematics. I have a course on number theory. Our professor told us a result that $U_n$ is cyclic iff $n=1,2,4,p^k,2p^k$ for $p$ odd prime. But I cannot find any easy proof of the result. This question is answered in stack exchange but the answer is without any explanation: For what $n$ is $U_n$ cyclic?

The answer says $U_{2^k}=\mathbb Z_2\times \mathbb Z_{2^{k-2}}$ for $k\geq 2$ and $U_{p^k}=\mathbb Z_{\phi(p^k)}$ which are beyond my understanding due to lack of explanation.

Can someone help me by providing a complete easy proof of this result using algebra?

Shaun
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Kishalay Sarkar
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1 Answers1

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I am not going to write out a detailed proof of a textbook result, but here are the essentials.

For $p=2$ check that $5 = 1+4$ has (multiplicative) order $2^{k-2}$ with $5^{2^{k-3}} \equiv 1+2^{k-1} \not\equiv -1 \bmod 2^k$, so $U(2^k) = \langle 5 \rangle \times \langle -1 \rangle \cong C_2 \times C_{2^{k-2}}$ for $k \ge 3$.

For $p$ odd, check that $1+p$ has order $p^{k-1}$. Choose $a$ such that $a$ has order $p-1$ mod $p$. Then the order of $a$ mod $p^k$ is $(p-1)p^t$ for some $t$, so $a^{p^t}$ has order $p-1$ and $U(p^k) = \langle 1+p \rangle \times \langle a^{p^t} \rangle \cong C_{p^{k-1}(p-1)} = C_{\phi(p^k)}$.

Derek Holt
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