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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3$U(m)$ 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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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 $$n*n + (-1)(n^2 - 1) = 1.$$
Cocopuffs
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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
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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. – bfhaha Mar 05 '16 at 06:10
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@bfhaha ig the reason may be, in a cyclic group of any finite order, no of elements of order $d$ is $\phi(d)$($\phi$ being the Euler's-Phi Function). But here no. of elements of order $2$ is $4$, whereas $\phi(2)=1$. This gives a contradiction. – Oogway Jan 18 '21 at 06:48
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@bfhaha theres a typo, it's $n^2-1$ , it's nice you invoked fundamental theorem of cyclic groups to show that, I guess the answerer himself wasn't realising $
$ is in fact $K4$, he just calculated some bunch of incongruent solution to show, more elements of order 2 exist. – M Desmond Oct 09 '21 at 12:26