dihedral group properties
1 Properties of Dihedral Groups
A group generated by two involutions is a dihedral group^{}. When the group is finite it is possible to show that the group has order $2n$ for some $n>0$ and takes the presentation^{}
$${D}_{2n}=\u27e8a,b{a}^{n}=1,{b}^{2}=1,{a}^{b}={a}^{1}\u27e9.$$ 
Remark 1.
Contemporary group theorists prefer ${D}_{\mathrm{2}\mathit{}n}$ over ${D}_{n}$ as the notation for the dihedral group of order $\mathrm{2}\mathit{}n$. Although this notation is overly explicit, it does help to resolve the ambiguity with the Lie type ${D}_{l}$ which corresponds to the orthogonal group^{} ${\mathrm{\Omega}}^{\mathrm{+}}\mathit{}\mathrm{(}\mathrm{2}\mathit{}l\mathrm{,}q\mathrm{)}$. However, introductory texts in algebra^{} still make use of the more appropriate ${D}_{n}$ notation to emphasize the connection to the symmetries^{} of a regular^{} $n$gon ($n$sided polygon^{}).
${D}_{2}\cong {\mathbb{Z}}_{2}$, ${D}_{4}\cong {\mathbb{Z}}_{2}\times {\mathbb{Z}}_{2}$ are the two abelian^{} examples of dihedral groups. They can be considered as dihedral groups of the respective order because they satisfy the relations^{}, though the geometric interpretations^{} are slightly modified. Often ${D}_{2}$ can be termed the symmetries of a line segment^{}, and the ${D}_{4}$ the symmetries of a nonsquare rectangle^{}, as the symmetry groups of each of these object is isomorphic^{} to ${D}_{2}$ and ${D}_{4}$ respectively. These exceptions cause problems for most theorems on dihedral groups so it is convenient to insist that $n>2$ for theorems.
Proposition 2.
$${D}_{2n}=\{{a}^{i}i\in {\mathbb{Z}}_{n}\}\bigsqcup \{{a}^{i}bi\in {\mathbb{Z}}_{n}\}$$ 
is a irredundant list of the elements of ${D}_{\mathrm{2}\mathit{}n}$. Moreover the conjugacy classes^{} of ${D}_{\mathrm{2}\mathit{}n}$ are $\mathrm{\{}{a}^{i}\mathrm{,}{a}^{\mathrm{}i}\mathrm{\}}$ for all $i\mathrm{\in}{\mathrm{Z}}_{n}$ and

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$2n$, $\{{a}^{2i}bi\in {\mathbb{Z}}_{n}\}$ and $\{{a}^{2i+1}bi\in {\mathbb{Z}}_{n}\}$

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$2\nmid n$, $\{{a}^{i}bi\in {\mathbb{Z}}_{n}\}$.
Consequently when $n\mathrm{>}\mathrm{2}$ the center of ${D}_{\mathrm{2}\mathit{}n}$ is $\mathrm{1}$ when $\mathrm{2}\mathrm{\nmid}n$ and $Z\mathit{}\mathrm{(}{D}_{\mathrm{2}\mathit{}n}\mathrm{)}\mathrm{=}\mathrm{\u27e8}{a}^{n\mathrm{/}\mathrm{2}}\mathrm{\u27e9}$ when $\mathrm{2}\mathrm{}n$. Furthermore ${C}_{n}\mathrm{:=}\mathrm{\u27e8}a\mathrm{\u27e9}$ is a characteristic subgroup of ${D}_{\mathrm{2}\mathit{}n}$, provided $n\mathrm{\ne}\mathrm{2}$.
Proof.
The conjugation^{} relation ${a}^{b}={a}^{1}$ allows us to place every element in the normal form ${a}^{i}{b}^{j}$. If ${a}^{i}{b}^{j}={a}^{k}{b}^{l}$ then ${a}^{ik}={b}^{lj}$. Yet $\u27e8a\u27e9\cap \u27e8b\u27e9=1$ so $ik\equiv 0\phantom{\rule{veryverythickmathspace}{0ex}}(modn)$ and $lj\equiv 0\phantom{\rule{veryverythickmathspace}{0ex}}(mod2)$. Thus we have and irredundant list as required.
For the conjugacy classes note that ${({a}^{i})}^{{a}^{j}b}={({a}^{i})}^{b}={a}^{i}$ so that these conjugacy classes are established. Next
$${({a}^{i}b)}^{{a}^{j}}={a}^{j+i}b{a}^{j}={a}^{j+i}{a}^{j}b={a}^{2j+i}b$$ 
for all $i\in \mathbb{Z}$. When $2\nmid n$ we have $(2,n)=1$ so $2$ is invertible modulo $n$. We let $j={2}^{1}(ki)$ for any $k\in \mathbb{Z}$ and we see that ${a}^{i}b$ is conjugate to any ${a}^{k}b$. However, when $2n$ we have a parity constraint that so far creates the two classes. We need to also verify conjugation by ${a}^{j}b$ does not fuse the two classes. Indeed
$${({a}^{i}b)}^{{a}^{j}b}={({a}^{2j+i}b)}^{b}=b{a}^{2j+i}b={a}^{2ji}b,$$ 
thus we retain two conjugacy classes amongst the reflections^{}.
Finally, the order of the elements $({a}^{i}b)$ is 2 – a fact used already. Thus the only cyclic subgroup of order $n$, when $n>2$, is ${C}_{n}$ and thus by its uniqueness it is characteristic. ∎
Proposition 3.
The maximal subgroups of ${D}_{\mathrm{2}\mathit{}n}$ are dihedral or cyclic. In particular, the unique maximal cyclic group is ${C}_{n}\mathrm{=}\mathrm{\u27e8}a\mathrm{\u27e9}$ and the maximal dihedral groups are those of the form $\mathrm{\u27e8}{a}^{n\mathrm{/}p}\mathrm{,}{a}^{i}\mathit{}b\mathrm{\u27e9}$ for primes $p$ dividing $n$.
We will prove this with a more general claim. First we pause to note that as a corollary to these two propositions^{} we can determine the entire lattice^{} of normal and characteristic subgroups of a dihedral group.
Corollary 4.
A proper subgroup^{} $H$ of ${D}_{\mathrm{2}\mathit{}n}$ is normal in ${D}_{\mathrm{2}\mathit{}n}$ if and only if $H\mathrm{\le}\mathrm{\u27e8}a\mathrm{\u27e9}$ or $\mathrm{2}\mathrm{}n$, and $H$ is one the following two maximal subgroups of index 2:
$${M}_{1}=\u27e8{a}^{2},b\u27e9,{M}_{2}=\u27e8{a}^{2},ab\u27e9.$$ 
The proper characteristic subgroups of ${D}_{\mathrm{2}\mathit{}n}$ are all the subgroups^{} of $\mathrm{\u27e8}a\mathrm{\u27e9}$.
Proof.
If $H$ is normal and contains an element of the form ${a}^{i}b$, then it contains the entire conjugacy class of ${a}^{i}b$. If $n$ is odd then all reflections are conjugate to ${a}^{i}b$ so $H$ contains all reflections of ${D}_{2n}$ and so $H$ is ${D}_{2n}$ as the relfections generate ${D}_{2n}$.
If instead $n$ is even then $H$ is forced only to contain one of the two conjugacy classes of reflections. If $i$ is even then $H$ contains $b$ and ${a}^{2}b$ so it contains ${a}^{2}$. If $i$ is odd then $H$ contains $ab$ and ${a}^{3}b$ so it contains ${a}^{2}=ab{a}^{3}b$ (note $n>3$ as $n>2$ and $2n$).
The two maximal subgroups of index $2$ which can exist when $n$ is even can be interchanged by an outer automorphism which maps $a\mapsto {a}^{1}$ and $b\mapsto ab$ so these two are not characterisitic. The subgroups of a characterisitic cyclic group are necessarily characteristic. ∎
Proposition 5.
Quotient groups^{} of dihedral groups are dihedral, and subgroups of dihedral groups are dihedral or cyclic.
Proof.
The homomorphic image^{} of a dihedral group has two generators^{} $\widehat{a}$ and $\widehat{b}$ which satisfy the conditions ${\widehat{a}}^{\widehat{b}}={\widehat{a}}^{1}$ and ${\widehat{a}}^{n}=1$ and ${\widehat{b}}^{2}=1$, therefore the image is a dihedral group.
For subgroups we proceed by induction^{}. When $n=1$ the result is clear. Now suppose that ${D}_{2n}$ has some proper subgroup $H$ that is not dihedral or cyclic. $H$ is contained in some maximal subgroup $M$ of ${D}_{2n}$. However the maximal subgroups of ${D}_{2n}$ are cyclic or dihedral so $H$ falls to the induction step for $M$ – together with the fact that subgroups of cyclic groups are cyclic. Thus $H$ must actually be dihedral or cyclic to avoid contradictions^{}. ∎
Proposition 6.
${D}_{n}$ is nilpotent if and only if $n\mathrm{=}{\mathrm{2}}^{i}$ for some $i\mathrm{\ge}\mathrm{0}$.
Proposition 7.
${D}_{2n}$ is solvable for all $n\mathrm{\ge}\mathrm{1}$.
Proof.
When $n=1$, ${D}_{2n}\cong {\mathbb{Z}}_{2}$ which is nilpotent and so also solvable. Now let $n>1$. Then ${D}_{2n}/\u27e8a\u27e9\cong {\mathbb{Z}}_{2}$ and $\u27e8a\u27e9\cong {\mathbb{Z}}_{n}$. Both ${\mathbb{Z}}_{n}$ and ${\mathbb{Z}}_{2}$ are nilpotent and so they are both solvable. As extensions^{} of solvable groups are solvable, ${D}_{2n}$ is solvable for all $n>0$. ∎
1.1 Automorphisms of ${D}_{2n}$
Theorem 8.
Let $n\mathrm{>}\mathrm{2}$. The automorphism group of ${D}_{\mathrm{2}\mathit{}n}$ is isomorphic to ${\mathrm{Z}}_{n}^{\mathrm{\times}}\mathrm{\u22c9}{\mathrm{Z}}_{n}$, with the canonical action of $\mathrm{1}\mathrm{:}{\mathrm{Z}}_{n}^{\mathrm{\times}}\mathrm{\to}\mathrm{Aut}\mathit{}{\mathrm{Z}}_{n}\mathrm{=}{\mathrm{Z}}_{n}^{\mathrm{\times}}$. Explicitly,
$$\mathrm{Aut}{D}_{2n}=\{{\gamma}_{s,t}s\in {\mathbb{Z}}_{n}^{\times},t\in {\mathbb{Z}}_{n}\}$$ 
with ${\gamma}_{s\mathrm{,}t}$ defined as
$$({a}^{i}){\gamma}_{s,t}={\alpha}^{is},({a}^{i}b){\gamma}_{s,t}={a}^{is+t}b.$$ 
Proof.
We apply the needleinthehaystack heuristic and search first to explain why these are the only possible forms for the automorphisms^{}. We will then prove all such are indeed automorphisms.
Given $\gamma \in \mathrm{Aut}{D}_{2n}$, we know $\u27e8a\u27e9$ is characteristic in ${D}_{2n}$ so $a\gamma ={a}^{s}$ for some $s\in {\mathbb{Z}}_{n}$. But $\gamma $ is invertible so indeed $(s,n)=1$ so that $s\in {\mathbb{Z}}_{n}^{\times}$. Next $b\gamma ={a}^{t}b$ as $b$ cannot be sent to $\u27e8a\u27e9$.
Now we claim $\gamma ={\gamma}_{s,t}$.
$$({a}^{i})\gamma ={a}^{is}=({a}^{i}){\gamma}_{s,t}$$ 
and
$$({a}^{i}b)\gamma ={a}^{is}{a}^{t}b={a}^{is+t}b=({a}^{i}b){\gamma}_{s,t}.$$ 
Now we must show all ${\gamma}_{s,t}$ are indeed homomorphisms^{} when $s\in {\mathbb{Z}}_{n}^{\times}$ and $t\in {\mathbb{Z}}_{n}$. First we note that $\gamma $ is welldefined as we have an irredundant listing of the elements. Next we verify the homomorphism cases.
$({a}^{i}{a}^{j}){\gamma}_{s,t}$  $=$  ${a}^{(i+j)s}={a}^{is}{a}^{js}=({a}^{i}){\gamma}_{s,t}({a}^{j}){\gamma}_{s,t}.$  
$({a}^{i}{a}^{j}b){\gamma}_{s,t}$  $=$  ${a}^{(i+j)s+t}b={a}^{is}({a}^{js+t}b)=({a}^{i}){\gamma}_{s,t}({a}^{i}b){\gamma}_{s,t}.$  
$({a}^{i}b{a}^{j}){\gamma}_{s,t}$  $=$  $({a}^{ij}b){\gamma}_{s,t}={a}^{(ij)s+t}b={a}^{is+t}b{a}^{js}=({a}^{i}b){\gamma}_{s,t}({a}^{j}){\gamma}_{s,t}.$  
$({a}^{i}b{a}^{j}b){\gamma}_{s,t}$  $=$  $({a}^{ij}){\gamma}_{s,t}={a}^{isjs}={a}^{is+ttis}$  
$=$  $({a}^{is+t}b)(b{a}^{tis})=({a}^{is+t}b)({a}^{js+t}b)=({a}^{i}b){\gamma}_{s,t}({a}^{j}b){\gamma}_{s,t}.$ 
So indeed ${\gamma}_{s,t}$ is a homomorphism.
Finally, we show the composition of two such maps both to identify the automorphism group and to show that each ${\gamma}_{s,t}$ is invertible.
$$({a}^{i}b){\gamma}_{s,t}{\gamma}_{u,v}=({a}^{is+t}b){\gamma}_{u,v}={a}^{isu+tu+v}b.$$ 
Hence, ${\gamma}_{s,t}{\gamma}_{u,v}={\gamma}_{su,tu+v}$. This agrees on ${a}^{i}$’s as well. This reveals the isomorphism desired: $\mathrm{Aut}{D}_{2n}\to {\mathbb{Z}}_{n}^{\times}\u22c9{\mathbb{Z}}_{n}$ by ${\gamma}_{s,t}\mapsto (s,t)$ where we see the multiplications agree as
$$(s,t)(u,v)=(su,tu+v).$$ 
In fact this demonstrates that the inverse^{} of ${\gamma}_{s,t}$ is simply ${\gamma}_{{s}^{1},t{s}^{}}$ and the identity map^{} is ${\gamma}_{1,0}$. ∎
Title  dihedral group properties 

Canonical name  DihedralGroupProperties 
Date of creation  20130322 16:06:35 
Last modified on  20130322 16:06:35 
Owner  Algeboy (12884) 
Last modified by  Algeboy (12884) 
Numerical id  10 
Author  Algeboy (12884) 
Entry type  Topic 
Classification  msc 20F55 
Related topic  GeneralizedQuaternionGroup 