Groups, G finite group from order 2p, p>2 prime and N is normal subgroup with index p in G , prove G cyclic i think i proved this question, but my proof isnt really elegant.
i assumed by contraditcion that there isnt exist an element from order 2p. then all the elements from lagrange are from order 2 or p.
Let N be {e,x}
First i proved that there cant be another group from order 2, which implies that all the elements are from order p.
let $g_1$ be an element from order p.
then i looked at the $G/N={N,g_1N,g_1^2N....,g_1^{p-1}N}$ and proved that all this cosets are distinct coset. 
Now because $g_1$ is from order p, we have another element g, which isnt belong to $<g_1>$ and  $N$ .
Now i proved that $gN$ isnt any of the coset that i mentiond above. so this is contratiction to the assumption that we have only p cosets.
I dont like this proof, seems to me not much elegant way to prove. Do you have another proof? 

Since $N$ has index $p$, $G/N$ has order $p$ and is therefore cyclic. Let $gN$ generate $G/N$.
In particular $g^pN=(gN)^p=N$. Suppose $G$ is not cyclic, then the order of $g$ must be $p$ (why?).
Since $N$ is normal in $G$, $g^{-1}xg=x$ (you might be able to say this without justification, but it could do with a justification).
Therefore the order of $gx$ is $2p$ (why?), so $G$ is cyclic. 
