Group of order $2p$ where $p$ is an odd prime is isomorphic to either integers mod $2p$ or dihedral group of order $2p$.

Question

$\newcommand{\Z}{\mathbb{Z}}$ Proposition: Let $p$ be an odd prime and $G$ a group of order $2p$. Then either $G \cong \Z/2p\Z$ or $G \cong D_{2p}$.

Attempt:

Well $G$ can be either cyclic or non cyclic, in the case where $G$ is cyclic it follows that $G \cong \Z/2p\Z$ as a finite cyclic group. So now assume that $G$ is not cyclic. By Cauchy's theorem there exist $r,s \in G$ for which $|r| = p$ and $|s| =2$ since both orders are primes diving $2p$. Note that the cyclic subgroup generated by $r$ must also have order $p$, and so $[G:\langle r \rangle ] = 2$, but his implies that $\langle r \rangle$ is normal in $G$. Because of this we get to form the quotient group as $G/\langle r \rangle$ and since there are only two cosets and one is $\langle r \rangle $ it follows that the other is $s \langle r \rangle$. Then because $G$ is partitioned into it's cosets it follows that $$ G = \{1,r,r^2,\dots,r^{p-1},s,sr,sr^2,\dots,sr^{p-1}\}, $$ and so this is almost exactly the dihedral group.

And then here I'm stuck. Is it enough to show that $rs = sr^{-1}$ because then $G$ satisfies all of the relations required of $D_{2p}$ and hence must be isomorphic?

How do I actually go about showing this relation holds? Moreover once this is shown is this enough to conclude? Or do I need to do more work in the case where $G$ is cyclic?

Also I can't use any of the Sylow theorems to solve this problem. Thanks in advance for any hints.

Answer 1

As per the recommendations of Arturo Magidin, I believe the proof can be finished as such.

We know that $srs^{-1} = r^j$ for some $1 \leq j \leq p-1$, this implies that $r = sr^js^{-1}$, but since conjugation by $s$ is an automorphism of $G$, and particularly a homomorphism it follows that $(sr^js^{-1}) = (srs^{-1})^j = r^{j^2}$. From this we conclude that $r = r^{j^2}$ and since $\langle r \rangle $ is a cyclic group we must have that $j^2 \equiv 1 \pmod p$. This implies that $j^2 = 1 +np$ for some $n \in \mathbb{Z}$ but in turn this implies that $$ j^2 - 1 = np \implies (j-1)(j+1) = np. $$ Because $p$ is a prime Euclid's lemma tells us that $p$ divides at least one of $(j-1)$ or $(j+1)$, or equivalently $j \equiv 1 \pmod p$ or $j \equiv -1 \pmod p$.

In the case where $j = 1\pmod p$ we have that $srs^{-1} = r^j = r$, and this implies that $sr = rs$, since these are commuting elements of coprime orders, we have that $|rs| = 2p$ and so $G$ is a cyclic group of order $2p$ and hence $G \cong \mathbb{Z}/2p\mathbb{Z}$.

In the case where $j \equiv -1\pmod p$ we have that $srs^{-1} = r^j = r^{-1}$ which implies $sr=r^{-1}s$, but then $G$ is exactly a presentation of the Dihedral group of order $2p$ (we already have $r^p =s^2 = 1$), therefore $G \cong D_{2p}$.

Note that $p$ cannote divide both $(j-1)$ and $(j+1)$ because then we would have $D_{2p} \cong \mathbb{Z}/2p\mathbb{Z}$ which is not possible since one is cyclic and the other isn't.