For finite group $G$ of order $n$ such that there exists a unique subgroup of order equal to each positive divisor of $n$. Is it necessarily cyclic?

Asked Aug 07 '15 at 13:53

Active Aug 07 '15 at 14:00

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Let $G$ be a finite group of order $n$. Suppose $k$ is a positive integer such that $k|n$, then there exists a unique subgroup of order $k$.

Is it necessarily cyclic?

If $G$ is Abelian, then I can prove it, but if $G$ is nonAbelian, then it is difficult to me to prove or disprove.

edited Aug 07 '15 at 14:00

asked Aug 07 '15 at 13:53

Parveen Chhikara

2

If G is the non-cyclic group of order 4 then it has a unique subgroup of order 4 (namely G itself) and this subgroup is non-cyclic. – lulu Aug 07 '15 at 13:55
I guess, you meant, there exists a unique subgroup of order $k$ for each $k|n$, else, of course every finite group of order $n$ has exactly one subgroup of order $n$. – Berci Aug 07 '15 at 13:56
Yes @Berci, I meant that. – Parveen Chhikara Aug 07 '15 at 14:01

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