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Assume that $G$ is a finite group such that for any positive integer $n$ dividing $|G|$, $G$ has one and only one subgroup $H$ with $|H|=n$. Is $G$ cyclic?

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    $\begingroup$ The order of an element of $g$ is a divisor of $|G|$. Try to count the number of elements of order $n$ for each divisor $n$ of $|G|$. $\endgroup$ – fkraiem Feb 17 '14 at 6:49
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Recall that $$\sum_{d \mid n} \phi(d) = n$$ where $\phi(\cdot) $ is the Euler totient function.

With your hypotheses, for every $d \mid n $ there are exactley $\phi(d) $ elements in $G$ of order $d$, i.e. the generators of the unique subgroup of order $d$.

So by the formula above there are also $\phi(n) $ elements of order $n \ $, i.e. the group is cyclic.

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Here is an alternative to the approach of WLOG (it uses some a bit more advanced group theoretic methods, but it uses a bit less number theoretic stuff. The proof by WLOG is definitely more neat, I just write this here to demonstrate a more group-theoretic proof):

Let $p$ be the smallest prime divisor of $|G|$ and write $|G| = p^mk$ with $\operatorname{gcd}(p,k) = 1$ (we can assume $G$ is not a $p$-group). Let $H\leq G$ with $|H| = p^m$. The assumption now implies that $H$ is normal, and the property described is inherited by the quotient $G/H$ which is thus cyclic by induction. Since the property is also inherited by $H$, $H$ is cyclic.
By the N/C theorem, we now get that $H$ is actually central in $G$, and since the quotient $G/H$ is cyclic, we must have that $G$ is abelian. But then we see that in fact, $G \cong H\times G/H$ and since these two factors are cyclic of coprime orders, we get that $G$ is cyclic.

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