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?

  • 1
    $\begingroup$ 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

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.


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.