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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 $n$?

  • Well, not $\infty$ but rather $|\Bbb Z|$. – Pedro Apr 14 '13 at 21:36
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    This is not a good candidate for induction. Simply give a direct proof by listing cosets. – Hagen von Eitzen Apr 14 '13 at 21:37
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    http://en.wikipedia.org/wiki/Euclidean_division – Martin Brandenburg Apr 14 '13 at 21:40
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    By the division algorithm, any $a \in \mathbb{Z}$ can be written (uniquely, in fact) as $a = q \cdot n + r$ for $(q,r) \in \mathbb{Z}^2$ and $0 \le r < n$. Therefore, $a \in r + n\mathbb{Z}$.

    A complete set of coset representatives is given by the possible remainders: ${ 0, 1, \ldots, n-1 }$.

     – Sammy Black Apr 14 '13 at 21:43
  • @PeterTamaroff: Is not $|\mathbb{Z}| = \infty$? –  Apr 14 '13 at 21:44
  • @HagenvonEitzen: I had originally done that, but I was just wondering if it is possible to prove by induction. –  Apr 14 '13 at 21:44
  • @PeterTamaroff: Is it that we should say that the order of $\mathbb{Z}$ is infinite as opposed to saying that it is infinity? –  Apr 14 '13 at 22:07
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    When speaking of the cardinality (the "number" of elements) of a set, what we're actually doing is establishing a bijection between the set and another (usually better known) set.

    There are different types of $\infty$, so @PeterTamaroff is just emphasizing that you can set up a bijection between the cosets $\mathbb{Z}/0\mathbb{Z}$ and $\mathbb{Z}$ itself.

     – Sammy Black Apr 15 '13 at 07:17

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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: $$ C_{k+1}=\bigcup_{i=0}^{k-1}\{i+k\mathbb{Z}\}\cup\{k+k\mathbb{Z}\}=\bigcup_{i=0}^{k}\{i+k\mathbb{Z}\}, \quad |C_{k+1}|=k+1. \qquad \blacksquare$$

scjorge
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