1

I have been reading: http://www.math.uconn.edu/~kconrad/blurbs/grouptheory/coset.pdf

Theorem 7.11 states: A finite group that has at most one subgroup of any size is cyclic.

I think this is a very crude formulation of theorem stated in this thread: If an Abelian group $G$ has order $n$ and at most one subgroup of order $d$ for all $d$ dividing $n$ then $G$ is cyclic

I believe the author of Theorem 7.11. forgot to explicitly state the order of each subgroup H has to the order of group G. Also, I don't really see, if the groups really have to be abelian or not, example (?)

As always: Any constructive hint, comment, answer or recommendation for further reading are appreciated. Thanks in advance.

Community
  • 1
Imago
  • 2,132

1 Answers1

1

You are correct, it should explicitly state any size that divides the order of $G$. However the statement is still true in general. Here is the sketch of proof. Since every subgroup of $G$ also has the same property, you may assume by induction that all subgroups of $G$ are cyclic. Also it is clear that all subgroups are normal in $G$. Since all Sylow $p$-subgroups of $G$ are normal we deduce that $G$ is nilpotent. Hence $G$ is a direct product of its Sylow subgroups which are all cyclic. So $G$ is cyclic.

Amin
  • 700
  • 1
    The group $\mathbb Z / 6\mathbb Z$ has exactly one subgroup for each divisor of $6$, but isn't a $p$-group. I think the condition that each subgroup is normal implies nilpotency, but not that it must be a $p$-group. – StefanH Oct 27 '15 at 10:52
  • @Stefan I meant without loss of generality we may assume that $G$ is a $p$-group. I edited my answer in a,I hope, more clear way. – Amin Oct 27 '15 at 18:54
  • Yes, I like your proof as it is way simpler than the counting arguments from the link. – StefanH Oct 27 '15 at 20:44