Exercise 9.2.1

Proof.

(a)

$\text{∙}$ Let $G$ be a cyclic group of order $n$ and let $g$ be a generator of $G$. If $f$ is a positive divisor of $n$, write $e=n∕f$, and $H=⟨g^e⟩$.

The order of $g$ is $n=\mathit{ef}$, hence the order of $g^e$ is $\frac{n}{n\wedge e}=\frac{n}{e}=f$, therefore the set $A=\{{(g^e)}^0,\cdots ,{(g^e)}^{f-1}\}\subset ⟨g^e⟩$ has $f$ distinct elements: $|A|=f$.

Conversely, if $h\in ⟨g^e⟩$, then $h={(g^e)}^k,k\in \mathbb{Z}$. The Euclidean division of $k$ by $f$ gives $k=\mathit{qf}+r,0\le r<f$, thus $h={(g^{\mathit{ef}})}^q{g}^{\mathit{er}}={(g^e)}^r,0\le r<f$, therefore $h\in A$.

Hence $H_f=⟨g^e⟩=A$ has order $f$.

$$|H_f|=|⟨g^e⟩|=f.$$

$\text{∙}$ Let $K$ be any subgroup of order $f$. We must prove that $K=H$.

The set $E$ of integers $m>0$ such that $g^m\in K$ is non empty, since $g^n=e\in K$. Set

$$k=\min <!--\; FUNCTION\; APPLICATION\; -->(E)=\min <!--\; FUNCTION\; APPLICATION\; -->\{m\in {\mathbb{N}}^{\text{∗}}|g^m\in K\},$$

so $k$ is the least positive integer such that $g^k\in K$. We show that $K=⟨g^k⟩$.

As $g^k\in K,⟨g^k⟩\subset K$.

Conversely, if $h\in K$, then $h$ is an element of $G$ of the form $h=g^l,l\in \mathbb{Z}$. The Euclidean division of $l$ by $k$ gives $l=\mathit{qk}+r,0\le r<k$.

Then $g^r=g^l{(g^k)}^{-q}=h{(g^k)}^{-q}\in K$ and $0\le r<k$. If $r$ was not zero, it would lie in $E$ and would be less than the minimum of $E$. This is a contradiction, so $r=0$, and $h=g^l={(g^k)}^q\in ⟨g^k⟩$. Therefore $K\subset ⟨g^k⟩$. Finally,

$$K=⟨g^k⟩.$$

We show first that $k\mid n$. Write $d=k\wedge n$. There exist integers $u,v$ such that $d=\mathit{uk}+\mathit{vn}$, therefore $g^d={(g^k)}^u{(g^n)}^v={(g^k)}^u\in K$, so $d\in E$, and $1\le d\le k$, therefore $d=k$ by definition of $k=\min <!--\; FUNCTION\; APPLICATION\; -->(E)$. So $k=k\wedge n$, hence $k\mid n$.

$K=⟨g^k⟩$ is cyclic, and its cardinality is the order of $g^k,k\mid n$, so

$$|K|=⟨g^k⟩=o(g^k)=\frac{n}{k},$$

by the first part of the proof.

By hypothesis the order of $K$ is $f$, so $f=|K|=n∕k$, and $k=n∕f=e$.

$$K=⟨g^e⟩=H.$$

Conclusion:

A cyclic group with generator $g$, of order $n=\mathit{ef}$, contains a unique subgroup of order $f$, written $H_f$, which is cyclic, generated by $g^e$.

(b)

Let $f,f^{\prime }$ be positive divisors of $p-1=|{(\mathbb{Z}∕\mathit{p\mathbb{Z}})}^{\text{∗}}|$, and let $g$ a generator of ${(\mathbb{Z}∕\mathit{p\mathbb{Z}})}^{\text{∗}}$. As in the text, write $H_f$ the unique subgroup of ${(\mathbb{Z}∕\mathit{p\mathbb{Z}})}^{\text{∗}}$ of order $f$.

If $H_f\subset H_f^{\prime }$, then $H_f$ is a subgroup of $H_f^{\prime }$. By Lagrange’s Theorem $|H_f|$ divides $|H_{f^{\prime }}|$, so $f\mid f^{\prime }$.

Conversely, if $f\mid f^{\prime }$, $f^{\prime }=\mathit{qf},q\in \mathbb{N}$. Moreover $H_f=⟨g^e⟩,H_{f^{\prime }}=⟨g^{e^{\prime }}⟩$, where $n=\mathit{ef}=e^{\prime }{f}^{\prime }$ by part (a). Therefore $e=e^{\prime }q$, and $g^e={(g^{e^{\prime }})}^q\in H_{f^{\prime }}$, hence $H_f=⟨g^e⟩\subset H_{f^{\prime }}$.

$$f\mid f^{\prime }⟺H_f\subset H_{f^{\prime }}.$$