Subgroup of Cyclic Group is Cyclic

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Theorem

Let $G$ be a cyclic group.

Let $H$ be a subgroup of $G$.

Then $H$ is cyclic.


Proof 1

Let $G$ be a cyclic group generated by $a$.

Let $H$ be a subgroup of $G$.

If $H = \set e$, then $H$ is a cyclic group subgroup generated by $e$.


Let $H \ne \set e$.

By definition of cyclic group, every element of $G$ has the form $a^n$.

Then as $H$ is a subgroup of $G$, $a^n \in H$ for some $n \in \Z$.

Let $m$ be the smallest (strictly) positive integer such that $a^m \in H$.

Consider an arbitrary element $b$ of $H$.

As $H$ is a subgroup of $G$, $b = a^n$ for some $n$.

By the Division Theorem, it is possible to find integers $q$ and $r$ such that $n = m q + r$ with $0 \le r < m$.

It follows that:

$a^n = a^{m q + r} = \paren {a^m}^q a^r$

and hence:

$a^r = \paren {a^m}^{-q} a^n$


Since $a^m \in H$ so is its inverse $\paren {a^m}^{-1}$

By Group Axiom $\text G 0$: Closure, so are all powers of its inverse.

Now $a^n$ and $\paren {a^m}^{-q}$ are both in $H$, thus so is their product $a^r$, again by Group Axiom $\text G 0$: Closure.

However:

$m$ was the smallest (strictly) positive integer such that $a^m \in H$

and:

$0 \le r < m$

Therefore it follows that:

$r = 0$

Therefore:

$n = q m$

and:

$b = a^n = \paren {a^m}^q$.

We conclude that any arbitrary element $b = a^n$ of $H$ is a power of $a^m$.

So, by definition, $H = \gen {a^m}$ is cyclic.

$\blacksquare$


Proof 2

Let $G$ be a cyclic group generated by $a$.


Finite Group

Let $G$ be finite.

By Bijection from Divisors to Subgroups of Cyclic Group there are exactly as many subgroups of $G$ as divisors of the order of $G$.

As each one of these is cyclic by Subgroup of Finite Cyclic Group is Determined by Order‎, the result follows.

$\Box$


Infinite Group

By Morphism from Integers to Group, an infinite cyclic group is isomorphic to $\struct {\Z, +}$.

So all we need to do is show that any subgroup of $\struct {\Z, +}$ is cyclic.

Suppose $H$ is a subgroup of $\struct {\Z, +}$.

From Subgroup of Integers is Ideal and Ring of Integers is Principal Ideal Domain, we have that:

$\exists m \in \Z_{> 0}: H = \ideal m$

where $\ideal m$ is the principal ideal of $\struct {\Z, +, \times}$ generated by $m$.

But $m$ is also a generator of the subgroup $\ideal m$ of $\struct {\Z, +}$, as:

$n \in \Z: n \circ m = n \cdot m \in \gen m$

Hence the result.

$\blacksquare$


Proof 3

Let $G$ be a cyclic group generated by $a$.

Let $H$ be a subgroup of $G$.

By Cyclic Group is Abelian, $G$ is abelian.

By Subgroup of Abelian Group is Normal, $H$ is normal in $G$.


Let $G / H$ be the quotient group of $G$ by $H$.

Let $q_H: G \to G / H$ be the quotient epimorphism from $G$ to $G / H$:

$\forall x \in G: \map {q_H} x = x H$

Then from Quotient Group Epimorphism is Epimorphism, $H$ is the kernel of $q_H$.

From Kernel of Homomorphism on Cyclic Group, $H$ is a cyclic group.

$\blacksquare$


Sources

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