prove that the group U(n^2 -1) is not cyclic

[2e4L]

Sorry if I formatted this thread incorrectly as its my first post ^^

Homework Statement

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

Homework Equations

The Attempt at a Solution

I've done a problem proving that U(2^n) is not cyclic when n >3, but I'm failing to make a parallel.

How does one find the order of (n^2 -1)? Is this information even needed to solve this problem?

 

[Deveno]

you might want to prove some easier things first:

suppose 8|k, that is:

k = 2^tm, where t ≥ 3.

then φ(k) = φ(2^t)φ(m), so

U(k) ≅ U(2^t) x U(m).

use this to show U(k) cannot be cyclic, as it has a non-cyclic subgroup.

now, consider even n and odd n for n²-1 separately.

 

[2e4L]

you might want to prove some easier things first:

suppose 8|k, that is:

k = 2^tm, where t ≥ 3.

then φ(k) = φ(2^t)φ(m), so

U(k) ≅ U(2^t) x U(m).

use this to show U(k) cannot be cyclic, as it has a non-cyclic subgroup.

now, consider even n and odd n for n²-1 separately.

I have learned that property of Euler's totient function, but not the one about U(ab) = U(a) x U(b). I've come across it online, but have yet to use it in class.

Is there another method?

 

[Deveno]

U(ab) isn't always isomorphic to U(a) x U(b).

for example U(4) is cyclic of order 2, but U(2) x U(2) has but a single element: (1,1).

it IS true, if a and b are co-prime.

this is actually a consequence of the chinese remainder theorem:

if gcd(m,n) = 1, then [a]_mn→([a]_m,[a]_n) is a group isomorphism of (Z_mn,+) with (Z_m,+) x (Z_n,+).

it's easy to check that [a]_mn→([a]_m,[a]_n) is a ring homomorphism as well, hence Z_mn and Z_m x Z_n have isomorphic groups of units (when...gcd(m,n) = 1. this is key).

i find it odd, that you wouldn't use this result, since it builds on the result of your previous problem.

 

[2e4L]

U(ab) isn't always isomorphic to U(a) x U(b).

for example U(4) is cyclic of order 2, but U(2) x U(2) has but a single element: (1,1).

it IS true, if a and b are co-prime.

this is actually a consequence of the chinese remainder theorem:

if gcd(m,n) = 1, then [a]_mn→([a]_m,[a]_n) is a group isomorphism of (Z_mn,+) with (Z_m,+) x (Z_n,+).

it's easy to check that [a]_mn→([a]_m,[a]_n) is a ring homomorphism as well, hence Z_mn and Z_m x Z_n have isomorphic groups of units (when...gcd(m,n) = 1. this is key).

i find it odd, that you wouldn't use this result, since it builds on the result of your previous problem.

It may be because it's still fairly early in the course. I learned the Chinese Remainder Theorem in a number theory class, but have yet to encounter it in my current one. We are going to study rings after the upcoming exam.

The previous problem was in a different chapter of the text I believe. This question is supposed to be a "review."