Groups of order $2p$ are the most concrete family among groups of order $pq$. When $p$ is an odd prime, the group is either cyclic or has the symmetry pattern of a regular $p$-gon. In this lesson, the focus keyword is dihedral groups and groups of order 2p. The discussion links Sylow theory with the familiar presentation of a dihedral group. Students often remember the relation $ba=a^{-1}b$ but forget that it comes from a conjugation action on a normal cyclic subgroup. :::definition[Dihedral Group] Let $n\geq 3$ be an integer. A group $D_n$ is called a dihedral group of degree $n$ if it is generated by elements $a$ and $b$ such that $$\begin{equation*} a^n=e,\qquad b^2=e,\qquad ba=a^{-1}b. \end{equation*}$$ ::: The element $a$ represents a rotation and the element $b$ represents a reflection. The relation $ba=a^{-1}b$ says that reflecting and then rotating is the same as rotating in the opposite direction and then reflecting. Algebraically, this relation controls every word in the generators. :::theorem Let $n\geq 3$. The dihedral group $D_n$ has order $2n$. ::: :::proof Given that $D_n=\langle a,b\rangle$ with $a^n=e$, $b^2=e$, and $ba=a^{-1}b$. To prove that $|D_n|=2n$. Using $ba=a^{-1}b$, every occurrence of $b$ can be moved to the right of a power of $a$. Since $b^2=e$, every element of $D_n$ can be written in one of the forms $$\begin{equation*} a^i\quad \text{or}\quad a^i b, \end{equation*}$$ where $0\leq i<n$. Thus $$\begin{equation*} D_n=\{e,a,a^2,\ldots,a^{n-1},b,ab,a^2b,\ldots,a^{n-1}b\}. \end{equation*}$$ The first $n$ elements are rotations and the last $n$ elements are reflections. These two types are distinct, and within each type the displayed elements are distinct because $o(a)=n$. Therefore $D_n$ has $2n$ elements. Hence $|D_n|=2n$. $\square$ ::: :::theorem Let $p$ be an odd prime. If $G$ is a group of order $2p$, then $G$ is cyclic or isomorphic to $D_p$. ::: :::proof Given that $p$ is an odd prime and $G$ is a group of order $2p$. To prove that $G$ is cyclic or isomorphic to $D_p$. Let $P$ be a Sylow $p$-subgroup of $G$. Since $n_p\equiv 1 \pmod p$ and $n_p\mid 2$, we have $n_p=1$. Therefore $P\trianglelefteq G$. Choose $a\in P$ with $P=\langle a\rangle$ and $o(a)=p$. By Cauchy's theorem, choose $b\in G$ with $o(b)=2$. Since $P$ is normal, conjugation by $b$ maps $P$ to itself. Thus $$\begin{equation*} bab^{-1}=a^r \end{equation*}$$ for some integer $r$. Since $b^2=e$, conjugation by $b$ has order dividing $2$. Therefore $$\begin{equation*} r^2\equiv 1 \pmod p. \end{equation*}$$ Thus $$\begin{equation*} r\equiv 1 \pmod p\quad \text{or}\quad r\equiv -1 \pmod p. \end{equation*}$$ If $r\equiv 1 \pmod p$, then $a$ and $b$ commute. Since $o(a)=p$, $o(b)=2$, and $\gcd(p,2)=1$, the element $ab$ has order $2p$. Therefore $G$ is cyclic. If $r\equiv -1 \pmod p$, then $$\begin{equation*} bab^{-1}=a^{-1}. \end{equation*}$$ Equivalently, $$\begin{equation*} ba=a^{-1}b. \end{equation*}$$ Thus $G$ has the defining relations of $D_p$. Hence $G$ is cyclic or $G\cong D_p$. $\square$ ::: This preview shows where the cyclic and dihedral cases come from. A group of order $2p$ has a normal cyclic subgroup $\langle a\rangle$ of order $p$. An element $b$ of order $2$ acts on $\langle a\rangle$ by conjugation, sending $a$ to $a^r$. The equation $r^2\equiv 1\pmod p$ leaves only the identity action and the inversion action. :::scientific-preview[Order Two p Action Explorer] ::: The two cases are two conjugation actions on the same normal cyclic subgroup. The identity action makes the factors commute, while the inversion action creates the dihedral relation. This calculator checks an action value $r$ modulo an odd prime $p$. The input is valid for a group of order $2p$ only when $r^2\equiv 1\pmod p$. The calculator identifies the cyclic case when $r\equiv 1$ and the dihedral case when $r\equiv -1$. Try $p=5$, $p=7$, and $p=11$. :::calculator[Cyclic or Dihedral Order Two p Calculator]