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Math Help - [SOLVED] Abstract Algebra: Integers Modulo n 2

  1. #1
    is up to his old tricks again! Jhevon's Avatar
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    [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 ). So feel free to respond anyway.


    Thanks guys and gals!
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Jhevon View Post
    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 View Post
    (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. My set theory is minimal basics only.

    -Dan
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  3. #3
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    Quote Originally Posted by Jhevon View Post
    (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]).
    (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.

    (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.
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