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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 $n^2-1=(n-1)(n+1)$. – lhf Aug 14 '12 at 16:24

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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+1 Nice!${}{}{}$ – Jyrki Lahtonen Aug 14 '12 at 16:34
Thanks so much for your help. I 've already known about primitive root theorem, but I really want a simple specific for this exercise. – ducquang98 Aug 14 '12 at 16:50
+1 Short, simple and nice indeed. – DonAntonio Aug 14 '12 at 17:38
@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. – bfhaha Mar 5 at 6:10

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