# Group of order $255$ is cyclic [duplicate]

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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 LahtonenJun 3 '17 at 19:55

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• That is certainly one correct way to prove the result. – Omnomnomnom Dec 8 '15 at 12:47
• 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 $G$. – Quang Hoang Dec 8 '15 at 12:51
• And $G/P$ seems to be a subgroup of $G$ of order $15$, isn't it? – user289143 Dec 8 '15 at 12:53
• @user289143 $G/P$ is not a subgroup of $G$. – user228113 Dec 8 '15 at 13:02
• @G.Sassatelli so what is your strategy to prove it? – user289143 Dec 8 '15 at 13:19

## 1 Answer

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)