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$Dih (C_{n}) = C_{n} ⋊ C_{2}$ with inversion action has order $2 n$ and the dihedral relations

Statement

For $n \geq 1$ , the generalized dihedral group of the cyclic group $C_{n}$ is the semidirect product

$Dih (C_{n}) = C_{n} ⋊ C_{2},$

where the nonidentity element of $C_{2}$ acts on $C_{n}$ by inversion. It has order $2 n$ , and if $C_{n} = ⟨ r ⟩$ and $C_{2} = ⟨ s ⟩$ , then

$r^{n} = s^{2} = 1, s r s^{- 1} = r^{- 1},$

and every element has a unique form $r^{i}$ or $r^{i} s$ with $0 \leq i < n$ .

This group is written $D_{n}$ here and called the dihedral group of order $2 n$ ; the notation is fixed by this Statement rather than assumed. At the two degenerate values the group is abelian: $Dih (C_{1}) ≅ C_{2}$ , and inversion on $C_{2}$ is the identity, so $Dih (C_{2}) ≅ C_{2} \times C_{2}$ .

Facts & Assumptions

Given: An integer $n \geq 1$ , $C_{n} = ⟨ r ⟩$ , and $C_{2} = ⟨ s ⟩$ .

[L1]

The generalized dihedral group is the semidirect product by the inversion action ( The generalized dihedral group $Dih (A) = A ⋊ C_{2}$ for an abelian group $A$ ).

[L2]

In an external semidirect product, every element has a unique factorisation from the two canonical subgroups, and conjugation induces the defining action (The canonical copy of $N$ is normal, the canonical copy of $H$ is a complement, and conjugation induces the action).

[L3]

Integer powers in a group satisfy the usual addition and inverse laws (Exponent laws in a group: $g^{m + n} = g^{m} g^{n}$ and $(g^{m})^{n} = g^{m n}$ for all $m, n \in Z$ , and $(g h)^{n} = g^{n} h^{n}$ when $g$ and $h$ commute).

Proof

technique · direct

1.1

Specialising [L1] to $C_{n}$ gives $C_{n} ⋊ C_{2}$ . The factor relations $r^{n} = 1$ and $s^{2} = 1$ hold in the canonical subgroups.

L1L2

1.2

The conjugation formula in [L2] gives $s r s^{- 1} = r^{- 1}$ . By [L3], it follows that $s r^{i} = r^{- i} s$ for every integer $i$ .

L2L3

2.1

Unique factorisation from [L2] says every pair is represented uniquely by $r^{i} s^{j}$ with $0 \leq i < n$ and $j \in {0, 1}$ . Hence $Dih (C_{n})$ has $2 n$ elements, in the two asserted forms, with the standard dihedral multiplication, and $r, s$ generate it.

L2step 1.2

3.1

At $n = 1$ the group $C_{1}$ is trivial, so $Dih (C_{1}) = C_{1} ⋊ C_{2} ≅ C_{2}$ ; at $n = 2$ every element of $C_{2}$ is its own inverse, so the inversion action is the identity and the semidirect product is direct, giving $Dih (C_{2}) ≅ C_{2} \times C_{2}$ . Both are consistent with the order and normal-form claims of step 2.1.

step 2.1L1L2∎

Depends on

Used by

The reflection complement in C₃ rtimes C₂≅ S₃ is not normal Counterexample
S₃≅ C₃ rtimes C₂ via inversion Example
False: the kernel and quotient determine a group extension up to isomorphism False statement

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 62 results over 20 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

Keith Conrad, Semidirect Products (standard reference, not scraped)

Dih⁡(Cn)=Cn⋊C2Dih(Cn​)=Cn​⋊C2​ with inversion action has order 2n2n and the dihedral relations