# dihedral group properties

## 1 Properties of Dihedral Groups

###### Remark 1.

Contemporary group theorists prefer $D_{2n}$ over $D_{n}$ as the notation for the dihedral group of order $2n$. 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  $\Omega^{+}(2l,q)$. 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.
 $D_{2n}=\{a^{i}~{}|~{}i\in\mathbb{Z}_{n}\}\sqcup\{a^{i}b~{}|~{}i\in\mathbb{Z}_{% n}\}$

is a irredundant list of the elements of $D_{2n}$. Moreover the conjugacy classes   of $D_{2n}$ are $\{a^{i},a^{-i}\}$ for all $i\in\mathbb{Z}_{n}$ 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 $n>2$ the center of $D_{2n}$ is $1$ when $2\nmid n$ and $Z(D_{2n})=\langle a^{n/2}\rangle$ when $2|n$. Furthermore $C_{n}:=\langle a\rangle$ is a characteristic subgroup of $D_{2n}$, provided $n\neq 2$.

###### 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 $a^{i-k}=b^{l-j}$. Yet $\langle a\rangle\cap\langle b\rangle=1$ so $i-k\equiv 0\pmod{n}$ and $l-j\equiv 0\pmod{2}$. 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

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

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 that so far creates the two classes. We need to also verify conjugation by $a^{j}b$ does not fuse the two classes. Indeed

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

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 $C_{n}$ and thus by its uniqueness it is characteristic. ∎

###### Proposition 3.

The maximal subgroups of $D_{2n}$ are dihedral or cyclic. In particular, the unique maximal cyclic group is $C_{n}=\langle a\rangle$ and the maximal dihedral groups are those of the form $\langle a^{n/p},a^{i}b\rangle$ for primes $p$ dividing $n$.

###### Corollary 4.

A proper subgroup  $H$ of $D_{2n}$ is normal in $D_{2n}$ if and only if $H\leq\langle a\rangle$ or $2|n$, and $H$ is one the following two maximal subgroups of index 2:

 $M_{1}=\langle a^{2},b\rangle,\qquad M_{2}=\langle a^{2},ab\rangle.$

The proper characteristic subgroups of $D_{2n}$ are all the subgroups   of $\langle a\rangle$.

###### 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 $a^{2}=aba^{3}b$ (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. ∎

###### Proof.

The homomorphic image   of a dihedral group has two generators   $\hat{a}$ and $\hat{b}$ which satisfy the conditions $\hat{a}^{\hat{b}}=\hat{a}^{-1}$ and $\hat{a}^{n}=1$ and $\hat{b}^{2}=1$, 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 $n=2^{i}$ for some $i\geq 0$.

###### Proposition 7.

$D_{2n}$ is solvable for all $n\geq 1$.

###### Proof.

When $n=1$, $D_{2n}\cong\mathbb{Z}_{2}$ which is nilpotent and so also solvable. Now let $n>1$. Then $D_{2n}/\langle a\rangle\cong\mathbb{Z}_{2}$ and $\langle a\rangle\cong\mathbb{Z}_{n}$. 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 $n>2$. The automorphism group of $D_{2n}$ is isomorphic to $\mathbb{Z}_{n}^{\times}\ltimes\mathbb{Z}_{n}$, with the canonical action of $1:\mathbb{Z}_{n}^{\times}\rightarrow\operatorname{Aut}\mathbb{Z}_{n}=\mathbb{Z% }_{n}^{\times}$. Explicitly,

 $\operatorname{Aut}D_{2n}=\{\gamma_{s,t}~{}|~{}s\in\mathbb{Z}_{n}^{\times},t\in% \mathbb{Z}_{n}\}$

with $\gamma_{s,t}$ defined as

 $(a^{i})\gamma_{s,t}=\alpha^{is},\qquad(a^{i}b)\gamma_{s,t}=a^{is+t}b.$
###### Proof.

Given $\gamma\in\operatorname{Aut}D_{2n}$, we know $\langle a\rangle$ 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 $\langle a\rangle$.

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.

 $\displaystyle(a^{i}a^{j})\gamma_{s,t}$ $\displaystyle=$ $\displaystyle a^{(i+j)s}=a^{is}a^{js}=(a^{i})\gamma_{s,t}(a^{j})\gamma_{s,t}.$ $\displaystyle(a^{i}a^{j}b)\gamma_{s,t}$ $\displaystyle=$ $\displaystyle a^{(i+j)s+t}b=a^{is}(a^{js+t}b)=(a^{i})\gamma_{s,t}(a^{i}b)% \gamma_{s,t}.$ $\displaystyle(a^{i}ba^{j})\gamma_{s,t}$ $\displaystyle=$ $\displaystyle(a^{i-j}b)\gamma_{s,t}=a^{(i-j)s+t}b=a^{is+t}ba^{js}=(a^{i}b)% \gamma_{s,t}(a^{j})\gamma_{s,t}.$ $\displaystyle(a^{i}ba^{j}b)\gamma_{s,t}$ $\displaystyle=$ $\displaystyle(a^{i-j})\gamma_{s,t}=a^{is-js}=a^{is+t-t-is}$ $\displaystyle=$ $\displaystyle(a^{is+t}b)(ba^{-t-is})=(a^{is+t}b)(a^{js+t}b)=(a^{i}b)\gamma_{s,% t}(a^{j}b)\gamma_{s,t}.$

So indeed $\gamma_{s,t}$ is a homomorphism.

Finally, we show the composition of two such maps both to 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: $\operatorname{Aut}D_{2n}\rightarrow\mathbb{Z}_{n}^{\times}\ltimes\mathbb{Z}_{n}$ 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 $\gamma_{s^{-1},-ts^{-}}$ and the identity map  is $\gamma_{1,0}$. ∎

Title dihedral group properties DihedralGroupProperties 2013-03-22 16:06:35 2013-03-22 16:06:35 Algeboy (12884) Algeboy (12884) 10 Algeboy (12884) Topic msc 20F55 GeneralizedQuaternionGroup