Converse Lagrange Theorem for groups with cyclic Sylow subgroups

Question

I try to solve the following problem ( Exercise 5c.4 from M.Isaacs Finite Group Theory).

Suppose that all Sylow subgroups of finite group $G$ are cyclic. Show that for every divisor $m$ of $|G|$ there exists a subgroup of order $m$ and show that every two subgroups of order $m$ are conjugate in $G$.

I prove the existence part of this problem by induction on $|G|$ using the following facts

1. $G'$ and $G/G'$ are cyclic groups of coprime orders.
2. $G$ is solvable

Also I have used the Shur-Zassenhaus theorem. But I don't know how to prove the conjugacy part. I tried to also reduce it to Schur-Zassenhaus theorem, but my attempts failed. So, I will be gratefull for ideas and hints.

Answer 1

Let $A$ and $B$ be subgroups of the same order $mn$, where $m$ and $n$ are divisors of $|G'|$ and $|G/G'|$ respectively. Then $A \cap G' = B \cap G'$ is the unique subgroup of $G'$ of order $m$, and it is normal in $G$.

Now, by Schur-Zassenhaus, $A \cap B'$ has complements $C$ and $D$ in $A$ and $B$, respectively. Since $G/G'$ is cyclic, we must have $G'C = G'D$ (its image in $G/G'$ is the unique subgroup of order $n$), so by Schur-Zassenhaus, $C$ and $D$ are conjugate in $G'C$, and hecne so are $A = (A \cap G')C$ and $B = (B \cap G')D$.

Answer 2

In general, every finite group $G$ with all Sylow subgroups being cyclic is supersolvable:

If every Sylow's subgroup is cyclic then $G$ is supersolvable.

It is well known that every supersolvable group is a CLT-group, i.e., where the converse of Lagrange's Theorem holds:

Complete classification of the groups for which converse of Lagrange's Theorem holds

Answer 3

@Derek Holt 's answer is completely fine, I will try to supply a proof with does not use Schur Zassenhous theorem's conjugacy part. Instead, we use Sylow theorems. Let $H,K$ be subgroups of order $m$ of $G$. We proceed by induction on order of $G$. Let $p$ be the largest prime dividing $m$. (So that Sylow $p$-subgroup of $H$ and $K$ are normal.)

Let $P\in Syl_p(G)$ such that $P\cap H\in Syl_p(H)$. There exists $g\in G$ such that $P^g\cap K\in Syl_p(K)$. Then $(P\cap H)^g$ and $P^g\cap K$ are subgroups of $P^g$ with eqaul order. Since $P$ is cyclic,

$$P^g\cap K=(P\cap H)^g.$$

Thus, $(P\cap H)^g\leq H^g\cap K$ and $(P\cap H)^g\lhd H^g$ and $(P\cap H)^g\lhd K$.

If $N_G((P\cap H)^g)<G$, then induction applied to $N_G((P\cap H)^g)$, $H^g$ and $K$ are conjugate and we are done.

We obtain that $(P\cap H)^g\lhd G$. Then induction apllied to $G/(P\cap H)^g$, $\overline K$ and $\overline H^g$ are conjugate in $\overline G$. It follows that they are indeed conjugate in $G$.