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A finite cyclic group has exactly one subgroup of each order dividing its own

Statement

Let $G = ⟨ g ⟩$ be a cyclic group of finite order $n \geq 1$ (The subgroup $⟨ S ⟩$ generated by a subset, the cyclic subgroup $⟨ g ⟩$ , and cyclic groups). For every positive divisor $d$ of $n$ (Divisibility in $Z$ : $d ∣ a$ when $a = d q$ for some integer $q$ ):

$⟨ g^{n / d} ⟩ is a subgroup of G of order d,$

it is the only subgroup of $G$ of order $d$ , and every subgroup of $G$ is of this form for exactly one such $d$ . Moreover, for positive divisors $d$ and $e$ of $n$ ,

$⟨ g^{n / d} ⟩ \subseteq ⟨ g^{n / e} ⟩ ⟺ d ∣ e .$

The two extremes are instances rather than exceptions: $d = 1$ gives the trivial subgroup ${e}$ and $d = n$ gives $G$ itself, and at $n = 1$ the only divisor is $d = 1$ , where $G$ is trivial.

Facts & Assumptions

Given: A cyclic group $G = ⟨ g ⟩$ whose underlying set is finite of order $n \geq 1$ ; divisibility of integers is that of Divisibility in $Z$ : $d ∣ a$ when $a = d q$ for some integer $q$ , and orders are those of The order $∣ G ∣$ of a finite group and the order $ord (g)$ of an element, with $ord (g) = \infty$ when no positive power of $g$ is the identity.

[A1]

$∣ G ∣ = n$ , and since $G = ⟨ g ⟩$ , [L1] gives $ord (g) = ∣ ⟨ g ⟩ ∣ = n$ .

[L1]

For an element $x$ of finite order $m$ in a group: $x^{k} = e$ if and only if $m ∣ k$ ; the powers $x^{0}, x^{1}, \dots, x^{m - 1}$ are pairwise distinct; and $⟨ x ⟩ = {x^{s} : s \in N, s < m}$ , so $⟨ x ⟩$ is finite with $∣ ⟨ x ⟩ ∣ = m = ord (x)$ (If $ord (g) = n$ then $g^{k} = e$ iff $k$ is an integer multiple of $n$ , the powers $g^{0}, \dots, g^{n - 1}$ are distinct, and $⟨ g ⟩$ has exactly $n$ elements; if $g$ has infinite order then $g^{j} = g^{k}$ only for $j = k$ ).

[L2]

$⟨ x ⟩ = {x^{k} : k \in Z}$ : the cyclic subgroup generated by $x$ is exactly the set of integer powers of $x$ ( $⟨ g ⟩ = {g^{n} : n \in Z}$ , and every cyclic group is abelian).

[L3]

For a finite group $G$ and $H \leq G$ one has $∣ G ∣ = [G : H] ∣ H ∣$ ; consequently $∣ H ∣$ divides $∣ G ∣$ (Lagrange's theorem: $∣ G ∣ = [G : H] ∣ H ∣$ for every subgroup $H$ of a finite group $G$ ).

Proof

technique · direct

1.1

Fix a positive divisor $d$ of $n$ and put $c : = n / d$ , an integer with $c \geq 1$ and $n = c d$ .

givenalgebra

2.1

The element $g^{c}$ has order exactly $d$ : first $(g^{c})^{d} = g^{c d} = g^{n} = e$ by [A1] and [L1]; and if $1 \leq j < d$ then $1 \leq c j < c d = n$ , so $n ∤ c j$ and hence $(g^{c})^{j} = g^{c j} \neq e$ by [L1]. Therefore $∣ ⟨ g^{c} ⟩ ∣ = d$ by [L1].

step 1.1A1L1algebra

2.2

The set of $x \in G$ with $x^{d} = e$ is exactly $⟨ g^{c} ⟩$ : writing $x = g^{m}$ with $m \in Z$ by [L2] and [A1], the condition $x^{d} = g^{m d} = e$ says $n ∣ m d$ by [L1], that is $c d ∣ m d$ , that is $c ∣ m$ ; and ${g^{m} : c ∣ m} = {(g^{c})^{k} : k \in Z} = ⟨ g^{c} ⟩$ by [L2].

step 1.1A1L1L2algebra

2.3

For positive divisors $d, e$ of $n$ with $d ∣ e$ , write $e = d f$ with $f \geq 1$ ; then $n / d = (n / e) f$ , so $g^{n / d} = (g^{n / e})^{f} \in ⟨ g^{n / e} ⟩$ by [L2], and therefore $⟨ g^{n / d} ⟩ \subseteq ⟨ g^{n / e} ⟩$ , the latter being a subgroup containing $g^{n / d}$ .

step 1.1L2algebra

3.1

If $H \leq G$ has $∣ H ∣ = d$ , then $H = ⟨ g^{c} ⟩$ : each $h \in H$ has $ord (h) = ∣ ⟨ h ⟩ ∣$ dividing $∣ H ∣ = d$ by [L1] and [L3], so $h^{d} = e$ by [L1] and hence $h \in ⟨ g^{c} ⟩$ by step 2.2; thus $H \subseteq ⟨ g^{c} ⟩$ , and both sets have exactly $d$ elements by step 2.1, so they are equal.

step 2.1step 2.2L1L3

4.1

Every subgroup $H \leq G$ has this form for exactly one positive divisor of $n$ : $H$ is a subset of the finite set $G$ , so $d : = ∣ H ∣$ is defined and divides $n$ by [L3] and [A1], and step 3.1 gives $H = ⟨ g^{n / d} ⟩$ ; the divisor is determined by $H$ , being its order.

step 2.1step 3.1A1L3

5.1

Conversely, if $⟨ g^{n / d} ⟩ \subseteq ⟨ g^{n / e} ⟩$ then $d ∣ e$ , since by step 2.1 the two subgroups have orders $d$ and $e$ and [L3] applied to the subgroup $⟨ g^{n / d} ⟩$ of $⟨ g^{n / e} ⟩$ makes $d$ divide $e$ . Together with steps 2.1, 3.1, 4.1 and 2.3 this proves every clause of the lemma.

step 2.1step 3.1step 2.3step 4.1L3∎

Remarks

What the divisor lattice buys. The clause that matters downstream is not existence but uniqueness: a subgroup of a finite cyclic group is pinned down by its order alone, so the Galois correspondence turns "subgroups of $Gal (F_{q^{n}} / F_{q})$ " into "divisors of $n$ " with nothing left to choose (The intermediate fields of $F_{q^{n}} / F_{q}$ are the $F_{q^{d}}$ , one for each positive divisor $d$ of $n$ ).
Where cyclicity is used. Uniqueness fails without it. In the Klein four-group there are three distinct subgroups of order two, and the argument breaks at step 2.2, where the solutions of $x^{2} = e$ form the whole group rather than a single cyclic subgroup.

Depends on

Used by

Dependency tree · two levels

39 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

J. S. Milne, Fields and Galois Theory, v5.10, Corollary 4.21 and its proof (standard reference, not scraped)
K. Conrad, Finite Fields (expository blurb), Theorem 5.2 (standard reference, not scraped)