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

  • 3
    $\begingroup$ $U(m)$ is cyclic iff $m$ is $2,4,p^k,2p^k$. Note that $n^2-1=(n-1)(n+1)$. $\endgroup$
    – lhf
    Aug 14, 2012 at 16:24

1 Answer 1


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 $$n*n + (-1)(n^2 - 1) = 1.$$

  • $\begingroup$ Thanks so much for your help. I 've already known about primitive root theorem, but I really want a simple specific for this exercise. $\endgroup$
    – ducquang98
    Aug 14, 2012 at 16:50
  • $\begingroup$ @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. $\endgroup$
    – bfhaha
    Mar 5, 2016 at 6:10
  • $\begingroup$ @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. $\endgroup$
    – Oogway
    Jan 18, 2021 at 6:48
  • $\begingroup$ @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 $<n>$ is in fact $K4$, he just calculated some bunch of incongruent solution to show, more elements of order 2 exist. $\endgroup$
    – M Desmond
    Oct 9, 2021 at 12:26

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