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Let $G$ a group and its order is $255$. Prove that $G$ is cyclic.

I easily demonstrated that the group has only one $17$-Sylow subgroup $P$ that is normal in $G$ and it's cyclic since it is of a prime order. Then $G/P$ is also cyclic since a group of order $15$ is cyclic. Then $G$ can be seen as $G=P(G/P)$ since the orders are coprime and then $G$ is cyclic. Is it correct?


marked as duplicate by Jyrki Lahtonen Jun 3 '17 at 19:55

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  • $\begingroup$ That is certainly one correct way to prove the result. $\endgroup$ – Omnomnomnom Dec 8 '15 at 12:47
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    $\begingroup$ Last step ($G=P(G/P)$) doesn't sound to me. You need to find a subgroup of $G$ of order $15$, not an image of $G$. $\endgroup$ – Quang Hoang Dec 8 '15 at 12:51
  • $\begingroup$ And $G/P$ seems to be a subgroup of $G$ of order $15$, isn't it? $\endgroup$ – user289143 Dec 8 '15 at 12:53
  • $\begingroup$ @user289143 $G/P$ is not a subgroup of $G$. $\endgroup$ – user228113 Dec 8 '15 at 13:02
  • $\begingroup$ @G.Sassatelli so what is your strategy to prove it? $\endgroup$ – user289143 Dec 8 '15 at 13:19

Yes I think it is correct. Because of the orders of all elements must divide to the order of the group and gcd(15,17)=1 you can prove that all the group is generated by only one element (by the same way you prove that a group of order 15 is cyclic)


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