dihedral group properties

1 Properties of Dihedral Groups

A group generated by two involutions is a dihedral group. Whenthe group is finite it is possible to show that the group hasorder $2n$ for some $n>0$ and takes the presentation

$$D_{2n}=⟨a,b|a^n=1,b^2=1,a^b=a^{-1}⟩.$$

Remark 1.

Contemporary group theorists prefer $D_{\mathrm{2}\mathit{}n}$ over $D_n$ as the notation for thedihedral 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 non-square 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.

$$D_{2n}=\{a^i|i\in {\mathbb{Z}}_n\}\bigsqcup \{a^{i}b|i\in {\mathbb{Z}}_n\}$$

is a irredundant list of the elements of $D_{\mathrm{2}\mathit{}n}$. Moreover the conjugacyclasses of $D_{\mathrm{2}\mathit{}n}$ are $\mathrm{\{}{a}^i\mathrm{,}{a}^{\mathrm{-}i}\mathrm{\}}$ for all $i\mathrm{\in }{\mathrm{Z}}_n$ and

• $2|n$, $\{a^{2i}b|i\in {\mathbb{Z}}_n\}$ and $\{a^{2i+1}b|i\in {\mathbb{Z}}_n\}$

• $2\nmid n$, $\{a^{i}b|i\in {\mathbb{Z}}_n\}$.

Consequently when $n\mathrm{>}\mathrm{2}$ the center of $D_{\mathrm{2}\mathit{}n}$ is $\mathrm{1}$ when $\mathrm{2}\mathrm{\nmid }n$ and$Z\mathit{}\mathrm{(}{D}_{\mathrm{2}\mathit{}n}\mathrm{)}\mathrm{=}\mathrm{⟨}{a}^{n\mathrm{/}\mathrm{2}}\mathrm{⟩}$ when $\mathrm{2}\mathrm{|}n$.Furthermore $C_n\mathrm{:=}\mathrm{⟨}a\mathrm{⟩}$ 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 elementin the normal form $a^i{b}^j$. If $a^i{b}^j=a^k{b}^l$ then$a^{i-k}=b^{l-j}$. Yet $⟨a⟩\cap ⟨b⟩=1$so $i-k\equiv 0(modn)$ and $l-j\equiv 0(mod2)$. Thus we have andirredundant list as required.

For the conjugacy classes note that ${(a^i)}^{a^{j}b}={(a^i)}^b=a^{-i}$ sothat 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 $j=2^{-1}(k-i)$ for any $k\in \mathbb{Z}$ and we see that $a^{i}b$ is conjugate to any $a^{k}b$. However, when $2|n$ we have a parity constraint thatso far creates the two classes. We need to also verify conjugation by$a^{j}b$ does not fuse the two classes. Indeed

$${(a^{i}b)}^{a^{j}b}={(a^{-2j+i}b)}^b=b{a}^{-2j+i}b=a^{2j-i}b,$$

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 $C_n$ andthus 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 $C_n\mathrm{=}\mathrm{⟨}a\mathrm{⟩}$ and themaximal dihedral groups are those of the form $\mathrm{⟨}{a}^{n\mathrm{/}p}\mathrm{,}{a}^i\mathit{}b\mathrm{⟩}$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 $H\mathrm{\le }\mathrm{⟨}a\mathrm{⟩}$ or $\mathrm{2}\mathrm{|}n$, and $H$ is one the following two maximal subgroupsof index 2:

$$M_1=⟨a^2,b⟩,M_2=⟨a^2,ab⟩.$$

The proper characteristic subgroups of $D_{\mathrm{2}\mathit{}n}$ are all the subgroups of $\mathrm{⟨}a\mathrm{⟩}$.

Proof.

If $H$ is normal and contains an element of the form $a^{i}b$, then it containsthe entire conjugacy class of $a^{i}b$. If $n$ is odd then all reflections areconjugate 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 $2|n$).

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 dihedralgroups 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 dihedralor cyclic. $H$ is contained in some maximal subgroup $M$ of $D_{2n}$. Howeverthe maximal subgroups of $D_{2n}$ are cyclic or dihedral so $H$ falls tothe induction step for $M$ – together with the fact that subgroups ofcyclic groups are cyclic. Thus $H$ must actually be dihedral or cyclicto 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 $D_{2n}/⟨a⟩\cong {\mathbb{Z}}_2$ and $⟨a⟩\cong {\mathbb{Z}}_n$. 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${\mathrm{Z}}_n^{\mathrm{\times }}\mathrm{⋉}{\mathrm{Z}}_n$, 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 needle-in-the-haystack heuristic and search first toexplain 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 $⟨a⟩$ is characteristicin $D_{2n}$ so $a\gamma =a^s$ for some $s\in {\mathbb{Z}}_n$. But $\gamma$ is invertibleso indeed $(s,n)=1$ so that $s\in {\mathbb{Z}}_n^{\times }$. Next $b\gamma =a^{t}b$as $b$ cannot be sent to $⟨a⟩$.

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 well-defined 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^{i-j}b){\gamma }_{s,t}=a^{(i-j)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^{i-j}){\gamma }_{s,t}=a^{is-js}=a^{is+t-t-is}$

$=$$(a^{is+t}b)(b{a}^{-t-is})=(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 identifythe 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: $\mathrm{Aut}{D}_{2n}\to {\mathbb{Z}}_n^{\times }⋉{\mathbb{Z}}_n$ 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 ${\gamma }_{1,0}$.∎