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I need to consider finite groups $G$ such that for any square-free number $d$ dividing the order of the group $G$, there exists a normal subgroup $H$ in $G$ such that either $H$ or $G/H$ has order $d$. Let us temporarily call such groups square-free-normal.

Problem. Does the class of square-free-normal finite groups coincide with some known class of finite groups?

Remark. It is known that a finite nilpotent groups contain an normal subgroup of any order dividing the order of the group, so nilpotent finite groups are square-free-normal. On the other hand, diherdal groups $D_{2p}$ with prime $p$ are square-free-normal but not nilpotent.

Question. What is the relation between square-free-normal finite groups and supersolvable finite groups? Is each supersolvable finite group square-free-normal?

Remark. Thank you for your comments, after which I arrive to the conclusion that square-free-normal supersolvable groups form a proper intermediate class between finite nilpotent groups and finite supersolvable groups. Im I right? Or this is known and has some other name? (I means "square-free-normal supersolvable")?

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    $\begingroup$ If $A$ is cyclic of order $n$ (let's say square free) and $B$ is any finite group of some order all of whose prime factors occur in $n$ then $A\times B$ is square-free-normal. I would assume that one might want to somehow exclude these degenerate examples. $\endgroup$
    – Tobias Kildetoft
    Feb 23, 2017 at 10:14
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    $\begingroup$ No, this case is Ok with me. In fact, my aim is to represent a "nice" (in my case square-free-normal supersolvable) group $G$ as the extension of a group $H$ and a group $K$ such that one of them is square-free and the other has order $k^2$ for some $k$ and contains a subgroup of order $k$. For this purpose I need square-free-normality + the CLT (the Converge Lagrange Theorem) property, possessed by supersolvable groups. In any case, thank you for the comment. $\endgroup$
    – Taras Banakh
    Feb 23, 2017 at 10:53
  • $\begingroup$ It sounds as though you mean the Converse Lagrange Theorem, which says that if $d$ is a positive integer with $d \mid |L|$ and $L$ is (for example) supersolvable, then $L$ has a subgroup of order $d$. $\endgroup$
    – DavidLHarden
    Feb 27, 2017 at 19:54

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Let $G=(C_7\rtimes C_3)\times(C_5\rtimes C_2)$.

Then $G$ is supersolvable but doesn't have a normal subgroup of order $7\times 2$ or $3\times 5$.

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  • $\begingroup$ Thank you. This answers my second question, but rather confirms the first one, name that the class of square-free-normal (supersolvable) groups rather does not coincide with anything known? $\endgroup$
    – Taras Banakh
    Feb 23, 2017 at 10:47

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