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 tostartthis, 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.