Coset/Examples/Subgroup of Infinite Cyclic Group
Examples of Cosets
Let $G = \gen a$ be an infinite cyclic group.
Let $s \in \Z_{>0}$ be a (strictly) positive integer.
Let $H$ be the subgroup of $G$ defined as:
- $H := \gen {a^s}$
Then a complete repetition-free list of the cosets of $H$ in $G$ is:
- $S = \set {H, aH, a^2 H, \ldots, a^{s - 1} H}$
Proof
First it is demonstrated that $S$ is complete.
Let $x \in G$ be arbitrary.
Then:
- $\exists n \in \Z: x = a^n$
By the Division Theorem:
- $n = q s + r$
for $q, r \in \Z$ such that $0 \le r \le s - 1$.
Thus:
\(\ds x\) | \(=\) | \(\ds a^{q s + r}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds a^r \paren {a^s}^q\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds a^r h\) | where $h \in H$ |
Thus:
- $x \in a^r H$
and as $0 \le r \le s - 1$ it follows that $a^r H \in S$.
Thus every $x \in G$ belongs to at least one of the cosets on $S$.
Aiming for a contradiction, suppose:
- $a^i H = a^j H$
where $0 \le i < j \le s - 1$.
Then from Sundry Coset Results:
- $\paren {a^i}^{-1} a^j \in H$
that is:
- $a^{j - i} \in H$
and so:
- $a^{j - i} = a^{k s}$
for some $k \in \Z$.
But $a$ is of infinite order.
So it follows from Subgroup Generated by Infinite Order Element is Infinite that:
- $j - i = k s$
But this contradicts our statement that $0 < j - i < s$.
Hence $S$ must be repetition-free.
$\blacksquare$
Sources
- 1978: Thomas A. Whitelaw: An Introduction to Abstract Algebra ... (previous) ... (next): $\S 42$. Another approach to cosets: Worked Example $3$