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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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With $n2 - 1$, do you mean all odd integers, or all integers of the form $n^2 - 1$?. The multiplicative group modulo a prime is cyclic, so it's not true for $U(2n - 1)$ Also, $U(2^2 - 1) = 3$, so you need $n > 2$ –  Lieven Aug 14 '12 at 16:12
What have you tried? If this is homework, tag it as such. –  lhf Aug 14 '12 at 16:22
$U(m)$ 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
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1 Answer

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
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