Coset/Examples/Subgroup of Infinite Cyclic Group

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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

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