Index of $n\mathbb{Z}$ in $\mathbb{Z}$. This is problem 1 of Ch.2, Section 6 in Artin's Algebra (Ed.1). It asks to determine the index $[\mathbb{Z} : n\mathbb{Z}]$. I have that if $|n|\geq 1$, then $[\mathbb{Z} : n\mathbb{Z}] = |n|$, but if $n=0$, then $[\mathbb{Z} : n\mathbb{Z}] = \infty$. Is this the case? Also, how would one go about proving the formula by induction for positive integers $n$?
 A: Consider the abelian groups $G=(\mathbb{Z},+)$ and $H=(n\mathbb{Z},+)$.
Answering your first question: yes, this is the case. If $|n|\geq1$, you have exactly $|n|$ cosets of $(n\mathbb{Z},+)$ in $G$ and these are:
$$0+H$$
$$1+H$$
$$...$$
$$(|n|-1)+H$$
Thus, $$[(\mathbb{Z},+):(n\mathbb{Z},+)]=|n|$$
For $n=0$, $0\mathbb{N}=\{0\}$ and the above list of cosets becomes infinite and countable. So, 
$$[(\mathbb{Z},+):(0\mathbb{Z},+)]=|\mathbb{N}|$$
as @PeterTamaroff noted.  
Addressing your second quiestion, let $n\in\mathbb{N}$.
We could make a proof by induction by defining a set $C_k$, whose elements are the $k$ cosets of the subgroup $H=(n\mathbb{Z},+)$ in $G=(\mathbb{Z},+)$: $$C_k:=\{\quad\bar{H}\subset G \,:\,\bar H=i+H \quad i=0,1,...,k-1\quad\} $$
Starting with $n=1$ we see that $\mathbb{Z}=n\mathbb{Z}$, so 
$$ C_1=\{0+1\mathbb{Z} \},\quad|C_1|=1.$$
For $n=2$ $$C_2=\{0+2\mathbb{Z},1+2\mathbb{Z}\}=C_1\cup\{1+2\mathbb{Z} \}, \quad |C_2|=2.$$ 
Supose then that for $n=k$, $$C_{k}=\bigcup_{i=0}^{k-2}\{i+k\mathbb{Z}\}\cup\{(k-1)+k\mathbb{Z}\}=\bigcup_{i=0}^{k-1}\{i+k\mathbb{Z}\}, \quad|C_{k}|=k.$$ Then for $n=k+1$ we get: 
$$ C_{k+1}=\bigcup_{i=0}^{k-1}\{i+k\mathbb{Z}\}\cup\{k+k\mathbb{Z}\}=\bigcup_{i=0}^{k}\{i+k\mathbb{Z}\}, \quad |C_{k+1}|=k+1. \qquad \blacksquare$$
