Alternative proof that $U(n^2-1)$ is not cyclic for $n>2$.

Question

I'm reading "Contemporary Abstract Algebra," by Gallian.

This is Exercise 4.84 ibid. and I want to solve it using only the tools available in the textbook so far. (A free copy of the book is available online.)

Notation: The group $$(\{a\in\Bbb Z_m\mid \gcd (a, m)=1\}, \times_m)$$ of units modulo $m$ under multiplication $\times_m$ (or concatenation) modulo $m$ is denoted $U(m)$.

The Question:

For every integer $n$ greater than $2$, prove that the group $U(n^2-1)$ is not cyclic.

Thoughts:

I'm aware that $n^2-1=(n-1)(n+1)$ as a difference of two squares.

That $U(n^2-1)$ is cyclic for $n=2$ is clear by direct computation.

External methods (e.g., ideas of proofs that rely on, say, anachronistic techniques):

• The result that $U(m)$ is cyclic iff $m$ is $2, 4, p^k, 2p^k$ for prime $p>2$ is not yet established in the text (I think).

• (What I suspect is) the lemma that if a group $G$ contains at least four distinct elements $x\in G$ such that $x^2=e$, then $G$ is not cyclic, is not clear to me; it's not established in the textbook so far and, yeah, I can sort of see why it's true (as the Klein four group is, intuitively, a (non-cyclic) subgroup of $G$). Please feel free to prove this lemma. It'd be sufficient, for me to understand the problem at hand.

• The Chinese Remainder Theorem is not established yet (but it is on page 347; I'm up to page 92).

• The theorem that $U(m)$ is cyclic iff $\varphi(m)=\lambda(m)$ is not established yet. (Here $\varphi$ is Euler's totient function and $\lambda$ is the Charmichael function.) In fact, I don't think it's mentioned at all (but I haven't looked very hard).

Please help :)

Answer 1

Using Theorem 4.4, one sees that if $2$ divides $\phi(n^2-1)$ (which is the order of $U(n^2-1)$, an even number) and this group is cyclic then there must be exactly $\phi(2) = 1$ elements of order 2.

The order of $n^2-2$ is $2$ (it is congruent to -1 modulo $n^2-1$) and the order of $n$ is clearly $2$ as well. This contradicts Theorem 4.4 if we assume that the group is cyclic.

Answer 2

Your second approach is right on: if $a=\pm 1, \pm n$, then $a^2 \equiv 1 \bmod n^2-1$.

The argument relies on this fact:

If $G$ is a cyclic group of order $m$ and $d$ divides $m$, then there is exactly one subgroup of $G$ of order $d$; it is the set of solutions of $x^d=1$.

For a proof, see here.