---- > [!definition] Definition. ([[dihedral group]]) > The [[group]] of symmetries of the regular $n-$gon is called the **dihedral group**. It contains $2n$ elements: $n$ rotations and $n$ reflections. Explicitly, if $x$ denotes the rotation by $\frac{2\pi}{n}$ in $D_{n}$ and $y$ any of the reflections in $D_{n}$, then $x^{n}=e, y^{2}=e, \text{ and } yxy^{-1}=x ^{-1},$ > and > $D_{n} = \{ e, x, \dots, x^{n-1}, y, yx, \dots, yx ^{n-1} \}.$ > \ > [[external semi-direct product|In general]], $D_{n} \cong C_{n} \rtimes C_{2}$, and products work as $x^{\ell}y=yx^{n-\ell}$. ^5c158d > [!basicexample] Example. (Lattice of Subgroups of the [[dihedral group]] $D_{4}$) > By taking powers of each element, we get the list > $\{e\},D_{4}, \langle x \rangle, \langle x^{2} \rangle, \langle y \rangle, \langle yx \rangle , \langle yx^{2} \rangle, \langle yx^{3} \rangle$, i.e., > $D_{4}, \{ e \}, \{e, x,x^{2}, x^{3} \}, \{ e, x^{2} \}, \{ e,y \}, \{ e, yx \},\{ e, yx^{2} \}, \{ e, yx^{3} \}$ of subgroups. There are two [[subgroup]]s not generated in this way; they are $\{ e, x^{2}, yx^{2}, y \} \and \{ e, x^{2}, yx, yx^{3} \}$. ![[CleanShot 2023-09-03 at [email protected]]] To show that these are the only [[subgroup]]s, let $S$ be a proper, nontrivial [[subgroup]] of $G$. Let $a \in S$ be nontrivial. Then by [[lagrange's theorem]] $|S|=2$ or $|S |=4$. \ Suppose $|S|=2$, $S=\{ e,a \}$. Then $S=\langle a \rangle$ and is hence listed above since we've enumerated all [[cyclic subgroup]]s of $D_{4}$ already. \ So assume $|S|=4$. $S$ must contain one of the [[cyclic subgroup]]s listed above; we'll proceed by cases to analyze what happens if we attempt to insert various elements into each one. > ## $\langle x \rangle \subset S$. In this case, since $|\langle x \rangle|=4$, we have $\langle x \rangle=S$.> > ## $\langle x^{2} \rangle \subset S$. *Insert $x$*: Then to keep $S$ closed under the [[binary operation|group operation]] we must also insert $(x x)x =x^{3}$. Because $|S|=4$, this immediately implies $S=\langle x \rangle$. \ *Insert $x^{3}$*: Then to keep $S$ closed under the [[binary operation|group operation]] we must also insert $(x^{3}x^{3})x^{3} =x$. Because $|S|=4$, this immediately implies $S=\langle x \rangle$. \ *Insert $yx$*: Then to keep $S$ closed under the [[binary operation|group operation]] we must also insert $(yx)x^{2}=yx^{3}$. Since $|S|=4$ this implies $S=\{ e,x^{2}, yx, yx^{3} \}$ as listed already. \ *Insert $yx^{2}$*: Then to keep $S$ closed under the [[group|group operation]] we must also insert $x^{2}(yx^{2})=y$. Since $|S|=4$ this implies $S=\{ e,x^{2}, yx^{2}, y \}$ as listed already. \ *Insert $yx^{3}$*: Then to keep $S$ closed under the [[group|group operation]] we must also insert $x^{2}(yx^{3})=yx$ (using that $x^{2}y=yx^{2}$). Since $|S|=4$ this implies $S=\{ e,x^{2}, yx, yx^{3} \}$ as listed already. \ *Insert $y$*: Then to keep $S$ closed under the [[group|group operation]] we must also insert $yx^{2}$. Since $|S|=4$ this implies $S=\{ e,x^{2}, yx^{2}, y \}$ as listed already. > ## $\langle y^{} \rangle \subset S$. *Insert $x$*: Then to keep $S$ closed under the [[binary operation|group operation]] we must also insert $x^{2}$, $x^{3}$, and $yx$. This yields $|S|>4$, a contradiction. So we cannot insert $x$. \ *Insert $x^{2}$*: Then to keep $S$ closed under the [[binary operation|group operation]] we must also insert $yx^{2}$. Since $|S|=4$ this implies $S=\{ e,x^{2}, yx^{2}, y \}$ as listed already. \ *Insert $x^{3}$*: Then to keep $S$ closed under the [[binary operation|group operation]] we must also insert $x^{6}=x^{2}$, $x^{9}=x$, and $yx^{3}$. This yields $|S|>4$, a contradiction. So we cannot insert $x^{3}$. \ *Insert $yx$*: Then to keep $S$ closed under the [[binary operation|group operation]] we must also insert $y(yx)=x$, and in turn must insert $x^{2}$. This yields $|S|>4$, a contradiction. So we cannot insert $yx$. \ *Insert $yx^{2}$*: Then to keep $S$ closed under the [[binary operation|group operation]] we must also insert $y(yx^{2})=x^{2}$. Since $|S|=4$ this implies $S=\{ e,x^{2}, yx^{2}, y \}$ as listed already. \ *Insert $yx^{3}$*: Then to keep $S$ closed under the [[binary operation|group operation]] we must also insert $y(yx^{3})=x^{3}$, and in turn $x^{3}x^{3}=x^{2}$. This yields $|S|>4$, a contradiction. So we cannot insert $yx^{3}$. > ## $\langle yx \rangle \subset S$. *Insert $x$*: Then to keep $S$ closed under the [[binary operation|group operation]] we must also insert $yx^{2}$, as well as $yx^{3}$. This yields $|S|>4$, a contradiction. So we cannot insert $x^{}$. \ *Insert $x^{2}$*: Then to keep $S$ closed under the [[binary operation|group operation]] we must also insert $yx^{3}$. Since $|S|=4$ this implies $S=\{ e,x^{2}, yx, yx^{3} \}$ as listed already. \ *Insert $x^{3}$*: Then to keep $S$ closed under the [[binary operation|group operation]] we must also insert $x^{3}yx=x(x^{2}y)x=xyx^{3}=yx^{2}$ as well as $x^{3}x^{3}=x^{2}$. This yields $|S|>4$, a contradiction. So we cannot insert $x^{3}$. \ *Insert $y$*: Then to keep $S$ closed under the [[binary operation|group operation]] we must also insert $yyx=x$, and in turn $x^{2}$. This yields $|S|>4$, a contradiction. So we cannot insert $y$. \ *Insert $yx^{2}$*: Then to keep $S$ closed under the [[binary operation|group operation]] we must also insert $yxyx^{2}=yxy^{-1}x^{2}= x ^{-1} x ^{2}=x$, and in turn $x^{2}$. This yields $|S|>4$, a contradiction. So we cannot insert $yx^{2}$. \ *Insert $yx^{3}$*: Then to keep $S$ closed under the [[binary operation|group operation]] we must also insert $yxyx^{3}=x^{2}$. Since $|S|=4$ this implies $S=\{ e,x^{2}, yx, yx^{3} \}$ as listed already. > ## $\langle yx^{2} \rangle \subset S$. *Insert $x$*: Then to keep $S$ closed under the [[binary operation|group operation]] we must also insert $x^{2}$ and $x^{3}$. This yields $|S|>4$, a contradiction. So we cannot insert $x^{}$. \ *Insert $x^{2}$*: Then to keep $S$ closed under the [[binary operation|group operation]] we must also insert $yx^{2}x^{2}=y$. Since $|S|=4$ this implies $S=\{ e,x^{2}, yx^{2}, y \}$ as listed already. \ *Insert $x^{3}$*: Then to keep $S$ closed under the [[binary operation|group operation]] we must also insert $yx^{2}x^{3}=yx$, and in turn $yxyx^{2}=x$. This yields $|S|>4$, a contradiction. So we cannot insert $x^{3}$. \ *Insert $y$*: Then to keep $S$ closed under the [[binary operation|group operation]] we must also insert $yyx^{2}=x^{2}.$ Since $|S|=4$ this implies $S=\{ e,x^{2}, yx^{2}, y \}$ as listed already. \ *Insert $yx$*: Then to keep $S$ closed under the [[binary operation|group operation]] we must also insert $yxyx^{2}=x$, and in turn $x^{2}$. This yields $|S|>4$, a contradiction. So we cannot insert $yx$. \ *Insert $yx^{3}$*: Then to keep $S$ closed under the [[binary operation|group operation]] we must also insert $yx^{2}yx^{3}=yyx^{2}x^{3}=x$, and in turn $x^{2}$. This yields $|S|>4$, a contradiction. So we cannot insert $yx^{3}$. > ## $\langle yx^{3} \rangle \subset S$. *Insert $x$*: Then to keep $S$ closed under the [[binary operation|group operation]] we must also insert $x^{2}$ and $x^{3}$. This yields $|S|>4$, a contradiction. So we cannot insert $x^{}$. \ *Insert $x^{2}$*: Then to keep $S$ closed under the [[binary operation|group operation]] we must also insert $yx^{3}x^{2}=yx$. Since $|S|=4$ this implies $S=\{ e,x^{2}, yx, yx^{3} \}$ as listed already. \ *Insert $x^{3}$*: Then to keep $S$ closed under the [[binary operation|group operation]] we must also insert $yx^{3}x^{3}=yx^{2}$, as well as $x^{3}x^{3}=x^{2}$. This yields $|S|>4$, a contradiction. So we cannot insert $x^{3}$. \ *Insert $y$*: Then to keep $S$ closed under the [[binary operation|group operation]] we must also insert $yyx^{3}=x^{3}$, and in turn $x^{3}x^{3}=x^{2}$. This yields $|S|>4$, a contradiction. So we cannot insert $y$. \ *Insert $yx$*: Then to keep $S$ closed under the [[binary operation|group operation]] we must also insert $yxyx^{3}=x^{2}$. Since $|S|=4$ this implies $S=\{ e,x^{2}, yx, yx^{3} \}$ as listed already. \ *Insert $yx^{2}$*: Then to keep $S$ closed under the [[binary operation|group operation]] we must also insert $yx^{2}yx^{3}=yyx^{2}x^{3}=x$, and in turn $x^{2}$. This yields $|S|>4$, a contradiction. So we cannot insert $yx^{2}$. ^d81188 > [!basicexample] Example. (Lattice of Subgroups of the Dihedral Group $D_{5}$) > Immediate [[subgroup]]s are $\{ e \}, D_{5}$. [[Cyclic subgroup]]s are $\langle x \rangle= \{ e,x,x^{2},x^{3},x^{4} \}, \langle y \rangle = \{ e,y \},$,$\langle yx \rangle= \{ e,yx \}, \langle yx^{2} \rangle=\{ e,yx^{2} \}, \langle yx^{3} \rangle = \{ e,yx^{3} \}, \langle yx^{4} \rangle=\{ e,yx^{4} \}$. > ![[CleanShot 2023-09-03 at [email protected]]] \ To show that these are the only [[subgroup]]s, let $S$ be a proper, nontrivial subgroup of $D_{5}$. By [[lagrange's theorem]], either $|S|=2$ or $|S|=5$. \ If $|S|=2$, $S=\{ e,a \}$, then $S=\langle a \rangle$ and is hence already listed since we've enumerated all [[cyclic subgroup]]s of $D_{5}$. \ So, suppose $|S|=5$ with $S \neq \langle x \rangle$. $S$ must contain only elements not in $\langle x \rangle$, since if it contained an element of $\langle x \rangle$ it would in turn contain $\langle x \rangle$ itself (yielding $|S|>|\langle x \rangle|=5$). That is, $S$ must be equal to one of $(D_{5} - \langle X \rangle - A_{i})\cup \{ e \}$, where $A:=\{ y,yx,yx^{2},yx^{3},yx^{4} \}$ is indexed by $i \in [6]$. Two cases: \ **1**. Suppose **$i=1$:** $S=\{e, yx,yx^{2},yx^{3},yx^{4} \}$. $(yx)yx^{2}=x$ must be in $S$ for $S$ to be a [[subgroup]], but by hypothesis $x \notin S$, so $S$ is not a [[subgroup]]. \ **2.** Else suppose **$i \neq 1$.** In this case $y\in S$ as well as $yx^{j}$ for $j \in [5]-i$. $y(yx^{j})=x^{j}$ must be in $S$ for $S$ to be a [[subgroup]], but by hypothesis $\langle x \rangle \cap S = \emptyset$, so $S$ is not a [[subgroup]]. \ We have shown that $(S \neq \langle x \rangle) \implies (S \text{ is not a subgroup of order 5})$. It follows that $\langle x \rangle$ is the only subgroup of $D_{5}$ with [[order of a group|order 5]]. > [!justification] > First we show that $x^{\ell}y=yx^{n-\ell}$. Start with the relation $yxy^{-1}=x ^{-1}$. Exponentiate both sides by $\ell$ to get $yx^{\ell}y^{-1}=x ^{-\ell}$. Then noting $y=y^{-1}$ left-multiply by $y$ on both sides to get $x^{\ell}y=yx^{-\ell}=yx^{n-\ell}$. ---- #### ---- #### References > [!backlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM [[]] > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ``` > [!frontlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM outgoing([[]]) > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ```