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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!

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2 Answers 2

up vote 6 down vote accepted

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.

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That is indeed a nice and short proof, thanks :) –  Serpahimz Jan 31 '13 at 14:45
    
More generally if $y^{-1}xy = x^t$ and $y^j = 1$, then $x = y^{-j}xy^j = x^{t^j}$, so $t^j \equiv 1 \mod{\operatorname{ord}(x)}$. This trick is sometimes useful for classifying finite groups of given order –  Mikko Korhonen Jan 31 '13 at 15:03

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 $2p$.

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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

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