Looking for a simple proof that groups of order $2p$ are up to isomorphism $\Bbb{Z}_{2p}$ and $D_p$ for prime $p>2$.

Question

I'm looking for a simple proof that up to isomorphism every group of order $2p$ ($p$ prime greater than two) is either $\mathbb{Z}_{2p}$ or $D_{p}$ (the Dihedral group of order $2p$).

I should note that by simple I mean short and elegant and not necessarily elementary. So feel free to use tools like Sylow Theorems, Cauchy Theorem and similar stuff.

Thanks a lot!

Answer 1

Since we are allowed to use Sylow, we can assume $G$ is generated by $x,y$ with $x^p=y^2=1$, where $\langle x \rangle \lhd G$, so $y^{-1}xy = x^t$ for some $t$ with $1 \le t \le p-1$. Then $x = y^{-2}xy^2 = x^{t^2}$, so $p$ divides $t^2-1 = (t-1)(t+1)$, hence $p$ divides $t-1$ or $t+1$ and the only possibilities are $t=1$ or $p-1$, giving the cyclic and dihedral groups.

Answer 2

Since the $2$-Sylow subgroup is cyclic, the group has a normal $2$-complement (corollary to Burnside's transfer theorem), which means that the $p$-Sylow subgroup is normal (or just use that any subgroup of index $2$ is normal). Thus, the group is a semidirect product of a cyclic group of order $p$ and one of order $2$. Since the Automorphism group of the cyclic group of order $p$ has a unique subgroup of order $2$, this means that there can only be one non-trivial such semidirect product, and since $D_p$ is such a semidirect product, it must be it. If the semidirect product is trivial, we of course get the cyclic group of order $2p$.

Answer 3

Use Cauchy Theorem

Cauchy's theorem — Let $G$ be a finite group and $p$ be a prime. If $p$ divides the order of $G$, then $G$ has an element of order $p$.

then you have an element $x\in G$ of order $2$ and another element $y\in G$ of order $p$. Now you have to show that $xy$ is of order $2p$
using commutativity we get:

$(xy)^2 = y^2$, and hence $ord(xy) \not| 2$

$(xy)^p = x$ , therefore $ord(xy) \not| p$

and $(xy)^{2p} = y^{2p} = e$, then $ord(xy) | 2p$

hence $1 <2<p<ord(xy) | 2p$
an then $ord(xy) = 2p$ because it doesn't equal to any divisor different of $2p$.

Answer 4

By Sylow-1 we can say that $\mathbb{Z}/(2)\simeq H<G$, $\mathbb{Z}/(p)\simeq K<G$, by Lagrange $H\cap K=\{e\}$ and it is easy ro see that $G=HK$.

And since $(G:K)=2$, $K\triangleleft G$.

So $G=K \rtimes_f H$ for some $f:H\to \operatorname{Aut}(K)$. It is well-known that $\operatorname{Aut}(\mathbb{Z}/(p))\simeq \mathbb{Z}/(p-1)$. So there are 2 such morphisms since $\gcd(2;p-1)=2$.

So, there are 1 or 2 groups of order $2p$, and we can name 2 of them: $\mathbb{Z}/(2p)$ and symmetries of a regular $n$-gon.