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Prove that $U(n^2−1)$ is not cyclic, where $U(m)$ is the multiplicative group of units of the integers modulo $m$.

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    $U(m)$ is cyclic iff $m$ is $2,4,p^k,2p^k$. Note that $n^2-1=(n-1)(n+1)$.2012-08-14

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For $n > 2$, $U(n^2-1)$ contains (at least) four distinct elements $x$ with $x^2 = 1$, namely $\pm 1, \pm n$ and this doesn't happen in cyclic groups. Note that $n$ is coprime to $n^2 - 1$ because $n*n + (-1)(n^2 - 1) = 1.$

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    @ducquang98 Cocopuffs doesn't use the primitive root theorem. He(or She) use the theorem: Any subgroup of a cyclic group must be a cyclic group as well. $S=\{1,-1,n,-n\}$ form a noncyclic subgroup of $U(2^n-1)$. ($S\cong \Bbb{Z}_2\oplus \Bbb{Z}_2$.) So $U(2^n-1)$ can't be a cyclic group.2016-03-05