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Let $G$ be a finite group of order $n$. If $G$ is cyclic, we then know that are subgroups are cyclic are are unique. If $G=\langle x\rangle$, and $d|n$, then $\langle x^d\rangle$ is the unique cyclic subgroup of order $\frac{n}{d}$.

However, suppose we know that the group $G$ has a unique subgroup of order $d$ for any $d$ such that $d|n$. What else can we say about $G$? Does it have to be cyclic? Can it be factorizable over subgroup $H$ and $K$?

Any help is greatly appreciated!

Scotty Vol
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1 Answers1

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This condition implies that the Sylow $p$-subgroup of $p$ is unique and cyclic. This implies that $G$ is the direct product of its Sylow subgroups, so is cyclic.

Angina Seng
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  • That's what I was thinking. I was hoping that there would be something that I was missing, but I wasn't 100% positive. Thank you. – Scotty Vol May 12 '17 at 16:20