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This question already has an answer here:

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!

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marked as duplicate by Derek Holt finite-groups May 12 '17 at 16:20

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ See the posted solutions to this question $\endgroup$ – lulu May 12 '17 at 16:18
  • $\begingroup$ Thanks. I don't know why I couldn't find that question earlier. $\endgroup$ – Scotty Vol May 12 '17 at 16:22
  • $\begingroup$ This question could be a duplicate target. It is a duplicate of quite a few other questions already. $\endgroup$ – Jyrki Lahtonen May 12 '17 at 16:22
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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.

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  • $\begingroup$ 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. $\endgroup$ – Scotty Vol May 12 '17 at 16:20

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