# Looking for a simple proof that groups of order $2p$ are up to isomorphism $\mathbb{Z}_{2p}$ and $D_{p}$ .

I'm looking for a simple proof that up to isomorphism every group of order 2p (p prime) 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!

-
your result is not true if you take p=2.So you should mention that p is an odd prime. – Ripan Saha Jul 15 '15 at 9:38
Even for $p=2$ it is true, with the convention that $D_2=C_2\times C_2$ is the Klein four group. – Dietrich Burde Mar 24 at 15:04

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.

-
Hi Derek, how is "$x = y^{-2}xy^2 = x^{t^2}$" true? – jstnchng Dec 7 '14 at 7:40
@jstn $x=e^{-1}xe=(y^2)^{-1}xy^2=y^{-1}(y^{-1}xy)y=y^{-1}x^t y=\underbrace{(y^{-1}xy)\cdot (y^{-1}xy)\cdots(y^{-1}xy)}_{\text{tea time}}=(x^t)^t=x^{t^2}$. – bfhaha Mar 21 at 9:34
Because $x$ has order $p$ and $y$ has order $2$ so the subgroup they generate has order divisible by $G$, and so it must equal $G$. – Derek Holt Mar 24 at 16:54
Yes you can assume that. – Derek Holt Mar 24 at 16:58
That is explained in the comment by bfhaha above. – Derek Holt Mar 24 at 18:24

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 $2p$.

-
You really don't need Burnside's transfer Theorem! A Sylow $p$-subgroup has index 2 and so must be normal. – Derek Holt Jan 31 '13 at 13:53
@DerekHolt I know (I also mentioned that simpler version). But as the proof can actually be done about as elegantly by elementary means, but the OP explicitly allowed advanced results, I could not resist using this approach (which can of course be used in much more general settings). – Tobias Kildetoft Jan 31 '13 at 13:55
For $p=2$ the trivial semidirect product is not the cyclic group, but it is $D_2$ in that case. But otherwise a nice proof. – Marc van Leeuwen Jan 31 '13 at 14:27
@MarcvanLeeuwen right, I assume $p\neq 2$ as otherwise the result is trivial. – Tobias Kildetoft Jan 31 '13 at 14:36