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).

Proposition 2   $$ D_{2n}=\{a^i~|~i\in \mathbb{Z}_n\}\sqcup \{a^ib ~|~i\in \mathbb{Z}_n\ $$ 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 $\{a^{2i+1}b ~|~ i\in \mathbb{Z}_n\}$
• $2\ndivides n$ , $\{a^ib~|~i\in \mathbb{Z}_n\}$ .

Consequently when $n>2$ the center of $D_{2n}$ is $1$ when $2\ndivides n$ 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 $\langle a\rangle \intersect \langle b\rangle=1$ 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 $$ (a^i b)^{a^j}=a^{-j+i}ba^j=a^{-j+i}a^{-j}b=a^{-2j+i} $$ for all $i\in \mathbb{Z}$ . When $2\ndivides n$ 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 $a^jb$ 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 $\langle a^{n/p}, a^ib\rangle$ 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 $a^ib$ 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 $a^2b$ so it contains $a^2$ . If $i$ is odd then $H$ contains $ab$ and $a^3b$ so it contains $a^2=aba^3b$ (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 dihedral groups are dihedral or cyclic.

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 $n>0$ .

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 $1:\mathbb{Z}_n^\times \rightarrow \Aut \mathbb{Z}_n = \mathbb{Z}_n^\times$ . Explicitly, $$ \Aut D_{2n}=\{\gamma_{s,t}~|~s\in \mathbb{Z}_n^\times, t\in \mathbb{Z}_n\ $$ 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. We apply the needle-in-the-haystack 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 $\gamma\in \Aut D_{2n}$ , 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 $$ (a^i b)\gamma=a^{is} a^tb=a^{is+t}b=(a^ib)\gamma_{s,t} $$ 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. \begin{eqnarray*} (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^ib 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^ib)\gamma_{s,t} (a^j)\gamma_{s,t}.\\ (a^iba^jb)\gamma_{s,t} & = & (a^{i-j})\gamma_{s,t} = a^{is-js}=a^{is +t -t -is}\\ & = & (a^{is+t}b)(ba^{-t-is})=(a^{is+t}b)(a^{js+t}b)=(a^ib)\gamma_{s,t}(a^jb)\gamma_{s,t}.\\ \end{eqnarray*}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. $$ (a^ib)\gamma_{s,t}\gamma_{u,v}=(a^{is+t}b)\gamma_{u,v}=a^{isu+tu+v}b $$ Hence, $\gamma_{s,t}\gamma_{u,v}=\gamma_{su,tu+v}$ . This agrees on $a^i$ 's as well. This reveals the isomorphism desired: $\Aut D_{2n}\rightarrow\mathbb{Z}_n^\times \ltimes \mathbb{Z}_n$ 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 $\gamma_{1,0}$ .