Let $G$ be a finite group and suppose that all Sylow subgroups of $G$ are cyclic. I have to prove that:
- If $m$ divides $|G|$ then there exists in $G$ a subgroup of order $m$.
- If $m$ divides $|G|$ then all subgroups of order $m$ are conjugate in $G$.
I want to prove both points by induction on $|G|$. If $p$ is the smallest prime that divides $|G|$ and $P$ is a $p$-Sylow of $G$, a corollary of Burnside's Theorem tells us that $P$ has a normal $p$-complement $N$ (and observe that the same argument works on $N$, this fact leads to a subnormal series with cylic factors, so $G$ is solvable). If $m$ is not a multiple of $p$ than $N$ has a subgroup of order $m$ by induction and two subgroups of order $m$ are $p'$-subgroups of $G$ that are contained in the unique Hall $p'$-subgroup $N$, so are conjugate by induction. Remains the case in which $p$ divides $m$. There is no loss of generality to assume that $p$ appears in $m$ with the same exponent $r$ wherewith appears in $|G|$. Consider the non empty set $\Omega =\{H\le N \mid |H|=m/p^r \}$. Then $P$ acts on $\Omega$ by conjugation; if each orbit was non trivial, than $|\Omega|$ would be a multiple of $p$, where by induction $N$ acts transitively on $\Omega$ and its order divides $|N|$ that is a $p'$-number. So there is a fixed point $H$, this meas that $P$ normalizes $H$ and $PH$ is a subgroup of the desired order. We have to prove now that all subgroups of order $m$ are conjugate, but I got stuck here and I don't know how to proceed.
I tried also to use the more advanced fact that the Fitting's group $F(G)$ is cyclic and $G/F(G)$ is cyclic, but I do not feel right this way.
Does anyone know a quicker way or how to produce the last part of my proof?
By induction,the intersections of the two subgroups $H$ and $K$ of order $m$ with $N$ are conjugate in $N$, so you by replacing one of them with a conjugate you can assume that $H \cap N = K \cap N = M$, say. Then $M$ is normal in $H$ and $K$ so we can assume that $M$ is normal in $G$. Now the result follows by applying induction to $G/M$.