Prove that if n > 2, then U_n has a subgroup of order 2.

Prove that if $\displaystyle n > 2$, then $\displaystyle \mathbb{U}_n$ has a subgroup of order $\displaystyle 2$.

There must exist a $\displaystyle g \in \mathbb{U}_n$ such that $\displaystyle g^2 \equiv 1 \pmod{n}$. How do I know there exists such an element in $\displaystyle \mathbb{U}_n$?

Re: Prove that if n > 2, then U_n has a subgroup of order 2.

how about n-1 (mod n)?

(there's one other element that always works, to make up "the rest of the subgroup". i'll leave it to you to discover. oh, and you'll want to show that gcd(n-1,n) = 1, as well).

Re: Prove that if n > 2, then U_n has a subgroup of order 2.

Thanks! Can I prove $\displaystyle gcd(n-1,n) = 1$ by the Euclidean Algorithm?

$\displaystyle n = (n-1) + 1$

Hence $\displaystyle gcd(n,n-1) = 1$.

Is the other element $\displaystyle 1$ that makes up the rest of the subgroup $\displaystyle \langle n-1 \rangle$?

I cannot find another generator for a subgroup of order $\displaystyle 2$ for $\displaystyle \mathbb{U}_5 = \{1,2,3,4\}$.

Re: Prove that if n > 2, then U_n has a subgroup of order 2.

well, of **course** 1 has the property that 1^{2} = 1 (mod n) (for any n, in fact).

4 has order 2 mod 5:

4^{2} = 16 = 1 (mod 5), so 4 is a generator (of a group of order 2).

if p is a prime, then U(p) will be cyclic (this is an involved proof, so i won't write it here), of order p-1.

as long as p > 2, then p will be odd, so U(p) (being cyclic) will have ONLY ONE element of order 2.

for a general n, there may be several elements of order 2 in U(n) (you get at least one for each prime factor of n).

Re: Prove that if n > 2, then U_n has a subgroup of order 2.

Thanks. I am having trouble understanding some of your points at the moment.

>> if p is a prime, then U(p) will be cyclic (this is an involved proof, so i won't write it here), of order p-1.

I found a theorem in my notes saying that if $\displaystyle n \geq 2$ then $\displaystyle \mathbb{U}_n$ is cyclic if and only if $\displaystyle n = 2,n=4,n=p^\alpha$ or $\displaystyle n = 2p^\alpha$, where $\displaystyle p$ is a prime other than $\displaystyle 2$ and $\displaystyle \alpha$ is a positive integer. The proof of this should cover the case when $\displaystyle p$ is prime.

I need to think more about the other two points. I will come back later.

>> as long as p > 2, then p will be odd, so U(p) (being cyclic) will have ONLY ONE element of order 2.

>> (you get at least one for each prime factor of n).