1
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – Robert Chamberlain Jun 9 '18 at 20:48
  • $\begingroup$ 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. $\endgroup$ – Moshe Levy Jun 9 '18 at 20:54
2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ thanks much better proof, please look at the comment above, just to check if my proof was ok, it was too long and annoying to prove the way i did. $\endgroup$ – Moshe Levy Jun 9 '18 at 21:02
  • $\begingroup$ Your comment completes your proof. It all looks correct. $\endgroup$ – Robert Chamberlain Jun 9 '18 at 21:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.