1. ## prove not cyclic

Hi All,

How to I Claim: For any interger n> 1, the group u(n(n+2)) is not cyclic?

Thanks.

2. Originally Posted by nguy0688
Hi All,

How to I Claim: For any interger n> 1, the group u(n(n+2)) is not cyclic?

Thanks.
If $\gcd(n,m) = 1$ then $\text{U}(\mathbb{Z}/nm\mathbb{Z}) \simeq \text{U}(\mathbb{Z}/n\mathbb{Z}) \times \text{U}(\mathbb{Z}/m\mathbb{Z})$ by the Chinese remainder theorem.

Now if $n$ is odd then $\gcd(n,n+2) =1$ and so $\text{U}(\mathbb{Z}/n(n+2)\mathbb{Z}) \simeq \text{U}(\mathbb{Z}/n\mathbb{Z}) \times \text{U}(\mathbb{Z}/(n+2)\mathbb{Z})$. Now this cannot be cyclic because the orders of those two groups are not relatively prime, since $\phi(x)$ is for $x>1$.

Similarly with $n$ odd, just write, $n(n+2) = 4(n/2)(n/2+1)$ and apply the argument.

3. ## but. . .

is there a way to prove this with out using the chinese remainder??? I am thinking about trying to prove it by showing that there is no element in u(n(n+2)) which generates the whole group. . . .but that's where i am stumped

4. Originally Posted by arsmath
is there a way to prove this with out using the chinese remainder??? I am thinking about trying to prove it by showing that there is no element in u(n(n+2)) which generates the whole group. . . .but that's where i am stumped
Maybe, I am just not sure how easy that method is. What is so bad with what I posted?

5. Originally Posted by ThePerfectHacker
Maybe, I am just not sure how easy that method is. What is so bad with what I posted?
nothing wrong with your solution. . .just me trying to avoid the Chinese remainder thm.