$\begingroup$

I think that this statement is false, but only in the trivial case. $\{ 0 \}$ is obviously a subgroup of $(\mathbb{Z},+)$ with infinite index.

In the case of a nontrivial subgroup of $(\mathbb{Z},+)$, intuition tells me that these subgroups will have finite index. How should I begin to prove this?

What I know:

Nontrivial subgroups of $\mathbb{Z}$ are of the for $n\mathbb{Z}$ for some $n\in\mathbb{N}$. These subgroups are infinite cyclic groups.

I read in another question posted here that I should try use the smallest positive element of $n\mathbb{Z}$ and Euclidean division to explicity describe all of $\mathbb{Z}$, but I'm not sure what that means, precisely.

$\endgroup$ 6

2 Answers

$\begingroup$

Hint: You want to show that $n\mathbb{Z}$ has finite index in $\mathbb{Z}$ for any $n>0$. To show this, try to show that any coset $x+n\mathbb{Z}$ is equal to one of the $n$ cosets $n\mathbb{Z}, 1+n\mathbb{Z}, 2+n\mathbb{Z},\dots,(n-1)+n\mathbb{Z}$.

$\endgroup$ $\begingroup$

Sketch of the proof: define $f: \mathbb{Z} \to \mathbb{Z}_n$ with $f(x) = [x]$. Now it is on you to prove that $f$ is a well-defined homomorphism onto $\mathbb{Z}_n$. Also prove that $ker(f)$ is isomorphic to $n\mathbb{Z}$.

$\endgroup$ 1

Your Answer

Sign up or log in

Sign up using Google Sign up using Facebook Sign up using Email and Password Submit

Post as a guest

Post Your Answer Discard

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy