Looking for a simple proof that groups of order $2p$ are up to isomorphism $\Bbb{Z}_{2p}$ and $D_p$ for prime $p>2$. 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!
 A: 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 $2p$.
A: If the group is Abelian then by the Classification of Finitely generated abelian groups we know that $G=\mathbb{Z}_{2p}$ is the only possibility (if $p$ is an odd prime).
If it is non Abelian:
By Cauchy's theorem one gets that there exits an element of order $p$ and order $2$. Call them $x$ and $y$ respectively, then all possible elements of the group are as follows:
$$\{1,x,x^2,...,x^{p-1},y,yx,yx^2,...,yx^{p-1},xy,x^2y,...x^{p-1}y\}$$ since the order of the group is $2p$, we get that $yx=x^{j}y$ for some $2\leq j\leq p-1$. ($j=1$ will force the group to be Abelian, so it not possible).
Then $yx^2=(yx)x=x^j(yx)=(x^{2j\bmod{p}})y\implies yx^k=(x^{kj\bmod{p}}y)$ by induction.
Then $y(yx)=x=y(x^jy)=(yx^j)y=x^{j^2 \bmod{p}}yy=x^{j^2 \bmod{p}}\implies x^{j^2 -1\bmod p}=1\implies j=\pm1\bmod p$.
Since the group is not Abelian we get that $yx=x^{p-1}y$ is the only reasonable relation possible, now consider the homorphism induced by $y\mapsto s$ and $r=x\mapsto r $. This is clearly an isomorphism to $D_{p}$.
Therefore only groups of order $2p$ are $\mathbb{Z_{2p}}$ or $D_{p}$.
A: 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.
A: 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.
A: 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 $2p$.
