# For what $n$ is $U_n$ cyclic?

When can we say a multiplicative group of integers modulo $n$, i.e., $U_n$ is cyclic?

$$U_n=\{a \in\mathbb Z_n \mid \gcd(a,n)=1 \}$$

I searched the internet but did not get a clear idea.

So $U_n$ is the group of units in $\mathbb{Z}/n\mathbb{Z}$.

Write the prime decomposition $$n=p_1^{\alpha_1}\cdots p_r^{\alpha_r}.$$

By the Chinese remainder theorem $$\mathbb{Z}/n\mathbb{Z}=\mathbb{Z}/p_1^{\alpha_1}\mathbb{Z}\times\ldots\times\mathbb{Z}/p_r^{\alpha_r}\mathbb{Z}$$ so $$U_n=U_{p_1^{\alpha_1}}\times\ldots\times U_{p_r^{\alpha_r}}.$$

For powers of $2$, we have $$U_2=\{0\}$$ and for $k\geq 2$ $$U_{2^k}=\mathbb{Z}/2\mathbb{Z}\times \mathbb{Z}/2^{k-2}\mathbb{Z}.$$

For odd primes $p$, $$U_{p^\alpha}=\mathbb{Z}/\phi(p^\alpha)\mathbb{Z}=\mathbb{Z}/p^{\alpha-1}(p-1)\mathbb{Z}.$$

So you see now that $U_n$ is cyclic if and only if $$n=2,4,p^\alpha,2p^{\alpha}$$ where $p$ is an odd prime.

Here is a reference.

• Why is it true that $U_{p^\k}=\mathbb{Z}/2\mathbb{Z}\times\mathbb{Z}/2^{k-2}\mathbb{Z}$? – Rasputin Jan 20 '17 at 20:03
• Julien, why doesn't the even prime work please? – BCLC Oct 17 '18 at 11:48
• By equal to symbol $=$ in proof. Did you mean isomorphic? – Akash Patalwanshi Jan 10 at 14:59

$U_n$ is cyclic iff $n$ is $2$, $4$, $p^k$, or $2p^k$, where $p$ is an odd prime.

The proof follows from the Chinese Remainder Theorem for rings and the fact that $C_m \times C_n$ is cyclic iff $(m,n)=1$ (here $C_n$ is the cyclic group of order $n$).

The hard part is proving that $U_p$ is cyclic and this follows from the fact that $\mathbb Z/p$ is a field and that $n = \sum_{d\mid n} \phi(d)$.

Any book on elementary number theory has a proof of this theorem. See for instance André Weil's Number theory for beginners, Leveque's Fundamentals of Number Theory, and Bolker's Elementary Number Theory.

Here "cyclic if and only if $\varphi(n)=\lambda(n)$" but there's no proof - the proof is elementary but very tricky.