# 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?

• Your last step $gN$ is not any of these cosets'' seems wrong. For example, $g=g_1x$ is not in $\langle g_1\rangle$ or $N$ but $gN=G_1N$. I'll give an outline of a neater proof as an answer. – Robert Chamberlain Jun 9 '18 at 20:48
• forgot to mention, proved that $N \subseteq Z(G)$ $gN=\{g,gx\}$ and $g1N=\{g_1,g_1x\}$ that means $g=g_1x$ which means $g^p=g_1^p*x^p$ which mean $e=x^p$ but p is odd, contradiction. – Moshe Levy Jun 9 '18 at 20:54

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.