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Thread: Determine if group / isomorphic

  1. #1
    Dec 2010

    Determine if group / isomorphic

    Let $\displaystyle n$ be positive integer and let $\displaystyle n\mathbb{Z} = {nm | m \in \mathbb{Z}}$.

    a. Show that $\displaystyle <n\mathbb{Z},+>$ is a group.

    b. Show that $\displaystyle <n\mathbb{Z},+> \simeq <\mathbb{Z},+>$

    For a ---

    Associativity -
    I defined binary operation $\displaystyle *$ as $\displaystyle a*b=na+ba$.

    $\displaystyle (a*b)*c=a*(b*c)$
    $\displaystyle (na+nb)*c=a*(nb+nc)$
    $\displaystyle n(na+nb)+nc=na+n(nb+nc)$
    $\displaystyle n^2a+n^2b+nc \neq na+n^2b+n^2c)$

    And therefore this is not associative and $\displaystyle <n\mathbb{Z},+>$ is not a group.

    Just for practice, I continued to the other axioms of a group.

    Identity element -

    I said there was no identity element, because for any value of e, $\displaystyle e*x$ and $\displaystyle x*e$ would leave a remaining x term paired with an n.

    $\displaystyle e*x = en+xn$
    $\displaystyle x*e = xn+en$

    When I try e=1, I get
    $\displaystyle e*x = n+xn$
    $\displaystyle x*e = xn+n$
    Which are both not x.

    When I try e=-x, I get
    $\displaystyle e*x = -xn+xn$
    $\displaystyle x*e = xn-xn$
    Which are both not x.

    So how do I prove that there is no identity, and if there is, how do I find it? I'm not not able to spot it right off the bat, am I doomed? Is there any sort of loose procedure?

    For the third axiom, the existence of an inverse, I said there was none because there is no identity element. I made this argument based on the equation in my book,
    $\displaystyle a*a'=a'*a=e$. Since e doesn't exist, I figured there can't be an inverse.

    For b ---

    I don't even know how to start this, so I'm just going to cross that bridge when I get to it.

    I know this is all completely wrong, any help is appreciated.
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  2. #2
    Member ModusPonens's Avatar
    Aug 2010

    Re: Determine if group / isomorphic

    You have to prove that nZ is a group under the given sum, not another operation you define. So

    i)associativity: since + is associative in Z it must also be associative in any nonempty subset of it, such as nZ.
    ii)zero: nZ is the set of elements kn, with k in Z, so 0n=0 also is in nZ
    ii)inverse: nZ is the set of elements kn, with k in Z. so if you add to kn, (-k)n, you'll get zero.

    b) Do you know anything about cyclic groups? It's realy handy to know cyclic groups when dealing with problems in Z, nZ or Z_n. Anyway, let f:Z -> nZ be such that f(k)=kn. Check that f is an homomorphism and check that it is injective and surjective, and you're done.
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  3. #3
    MHF Contributor

    Mar 2011

    Re: Determine if group / isomorphic

    the one thing ModusPonens did not verify above is that + is a valid binary operation on nZ (closure). fortunately, this is simple:

    suppose we have nk,nm in nZ (since nk = kn in Z we could have written this the other way around, but i'm just following the way the set is given in the problem).

    then nk+nm = n(k+m), and since k+m is in Z, n(k+m) is in nZ.

    the really interesting thing here is that the map kn-->n is an isomorphism. the set nZ is intuitively "smaller" than Z, it's only every multiple of n. so it seems counter-intuitive that it should be the same size as Z. but: both sets are infinite, indeed countably infinite, so it is easy to set up a 1-1 correspondence (bijection) that preserves the additive operation. we never "run out" of multiples of n to pair with each integer, even though we are "skipping" n numbers each time we do. this can only happen with infinite groups, finite groups are better-behaved.
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