Quotient Group of Infinite Cyclic Group by Subgroup
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Theorem
Let $C_n$ be the cyclic group of order $n$.
Then:
- $C_n \cong \dfrac {\struct {\Z, +} } {\struct {n \Z, +} } = \dfrac \Z {n \Z}$
where:
- $\Z$ is the additive group of integers
- $n \Z$ is the additive group of integer multiples
- $\Z / n \Z$ is the quotient group of $\Z$ by $n \Z$.
Thus, every cyclic group is isomorphic to one of:
- $\Z, \dfrac \Z \Z, \dfrac \Z {2 \Z}, \dfrac \Z {3 \Z}, \dfrac \Z {4 \Z}, \ldots$
Proof
Let $C_n = \gen {a: a^n = e_{C_n} }$, that is, let $a$ be a generator of $C_n$.
Let us define $\phi: \struct {\Z, +} \to C_n$ such that:
- $\forall k \in \Z: \map \phi k = a^k$
Then from the First Isomorphism Theorem:
- $\Img \phi = C_n = \struct {\Z, +} / \map \ker \phi$
We now need to show that $\map \ker \phi = n \Z$.
We have:
- $\map \ker \phi = \set {k \in \Z: a^k = e_{C_n} }$
Let $x \in \map \ker \phi$.
Then $a^x = e_{C_n}$ and thus:
- $n \divides x$
The result follows.
$\blacksquare$
Sources
- 1965: J.A. Green: Sets and Groups ... (previous) ... (next): $\S 7.4$. Kernel and image: Example $141$