# [SOLVED] Abstract Algebra: Integers Modulo n 2

• Feb 15th 2008, 01:42 PM
Jhevon
[SOLVED] Abstract Algebra: Integers Modulo n 2
Hallo,

Me again. Here's the last of four questions I have for today. (I'll probably asking some questions about Set Theory soon, but not today).

Again, this is a homework problem, so please, only hints

Problem:

(a) Prove that if $n$ is even, then exactly one nonidentity element of $\mathbb{Z}_n$ is its own inverse.

(b) Prove that if $n$ is odd, then no nonidentity element of $\mathbb{Z}_n$ is its own inverse.

(c) Prove that $[0] \oplus [1] \oplus \cdots \oplus [n - 1]$ equals either $[0]$ or $[n/2]$ in $\mathbb{Z}_n$

(d) What does part (c) imply about $0 + 1 + \cdots + (n - 1)$ modulo $n$?

Things that might come in handy:

Nothing really. Again, anyone who can help me will already know all they need to know by looking at the symbols and interpreting them.

What I've tried:

I honestly have not tried anything as yet. I don't think this is a particularly hard question. I remember being confused about something when I read it through the first time though, so I post it just in case, so as to not leave it for the last minute (which is when I usually do my homework).

I won't look at any hints given here until I've tried to do the problem myself (you can trust me :D). So feel free to respond anyway.

Thanks guys and gals!
• Feb 15th 2008, 05:09 PM
topsquark
Quote:

Originally Posted by Jhevon
Problem:

(a) Prove that if $n$ is even, then exactly one nonidentity element of $\mathbb{Z}_n$ is its own inverse.

Probably the best way to attack this one is to construct the element. Which do you think would be the likely candidate? If you can't think of which, Start with $\mathbb{Z}_4$ and work your way up. The pattern should be obvious.

Quote:

Originally Posted by Jhevon
(b) Prove that if $n$ is odd, then no nonidentity element of $\mathbb{Z}_n$ is its own inverse.

Easy once you've seen what happens in a).

You're on your own for the rest. (Nod) My set theory is minimal basics only.

-Dan
• Feb 16th 2008, 02:48 PM
ThePerfectHacker
Quote:

Originally Posted by Jhevon
(a) Prove that if $n$ is even, then exactly one nonidentity element of $\mathbb{Z}_n$ is its own inverse.

If $n$ is even then $n/2$ is an interger and $[n/2] + [n/2] = [n] = [0]$ so it is its own inverse. Now to prove uniqueness suppose that $[x]$ is a congruence class so that $[x]+[x]=[0]$ this means $[2x]=[0]$ and thus $2x\equiv 0(\bmod n)$. This means $n|2x \implies (n/2)|x$ thus $x = (n/2)k$ for an integer $k$. Now this implies for $k=0$ that $[0]$ has its own inverse, but we are considering non-identity elements and $k=1$ implies $[n/2]$ which is what we found. Thus, there are no other ones (just note if $k$ is even you end up with $[0]$ otherwise with $[n/2]$).
Quote:

(b) Prove that if $n$ is odd, then no nonidentity element of $\mathbb{Z}_n$ is its own inverse.
Suppose $[x]+[x] = [0]$ then $2x\equiv 0(\bmod n)$ so $n|2x$ since $\gcd(2,n)=1$ ( $n$ is odd) it means $n|x \implies x \in [0]$ thus only the identity element is its own inverse.

Quote:

(c) Prove that $[0] \oplus [1] \oplus \cdots \oplus [n - 1]$ equals either $[0]$ or $[n/2]$ in $\mathbb{Z}_n$

(d) What does part (c) imply about $0 + 1 + \cdots + (n - 1)$ modulo $n$?
Let $\{ a_1,a_2,...,a_n\}$ be a permutation of the intergers $\{ 0,1,2,...,n-1\}$ but not necessarily in that order. It has a special order to it. Then $0+1+...+(n-1) = a_1+a_2+...+a_n$ now the permutation has the special order is that $a_1,a_2$ are inverse $a_3,a_4$ are inverses .... Now if $n$ is odd then only one element is its own inverse i.e. 0, but everything else has a pair to it which cancels. So it means $a_1+a_2+...+a_n = 0$ if $n$ is odd. If $n$ is even everything except 0 has a mate except $n/2$ which has its self as its own mate which cancels out. Thus, $a_1+...+a_n \equiv n/2$.