Converse Lagrange Theorem for groups with cyclic Sylow subgroups

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.

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 $B = (B \cap G')D$.

• I see, this is nice. Commented Mar 25, 2018 at 11:22
• Thank you very much, Derek. I have one question. How it is easily to see that $|A\cap G'|=m$? For this, I have used induction on $|G|$ (that all subgroups of given order are conjugated), working in $MG'$, where $M$ is a subgroup of order $m$ in $A$ and then used induction hypothesis. Maybe it can be shown directly? Commented Mar 25, 2018 at 12:08
• $G/A \cap G'$ has order dividing $|G/G'|$, which is coprime to $|G'|$, so we must have $|A \cap G'| = m$, $|G/|A \cap G'| = n$. Commented Mar 25, 2018 at 13:54
• maybe you want to write " $A/A\cap G'$ has order dividing $G/G'$"? Commented Mar 25, 2018 at 14:25
• Yes maybe I do :) Commented Mar 25, 2018 at 15:21

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

• But the question was mainly about the conjugacy of subgroups of the same order. have you answered that? Commented Mar 25, 2018 at 11:08
• No, I only wanted to point out why "in general" the statement is true. And it says " So, I will be grateful for ideas and hints" - in general, I suppose. Commented Mar 25, 2018 at 11:09
• @DietrichBurde, Thank you for refference. Commented Mar 25, 2018 at 13:10

@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 $G$.

• Thank you, mesel! Very interesting solution. I have one question. Why the conjugacy of factor groups implies the conjugacy of groups in the last part of your solution? Commented Mar 25, 2018 at 14:21
• @MikhailGoltvanitsa: Let say $\overline H^{gs}=\overline K$. Then $H^{gs}/X=K/X$ where $X=(P\cap H)^g$. By correspondence theorem, $H^{gs}=K$. (The key fact is that the normal group is contained in both group $K$ and $H^g$) Commented Mar 25, 2018 at 14:27
• Oh..really, thank you very much. Commented Mar 25, 2018 at 14:30
• @MikhailGoltvanitsa: You are welcome. Commented Mar 25, 2018 at 14:32