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.

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.

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.