# Some properties of a finite group with all Sylow subgroups that are cyclic

I consider a finite group $G$ such that all its Sylow's subgroups are cyclic. I suppose that $|G|=p_1^{k_1}...p_n^{k_n}$ with $p_1<...<p_n$ distinct primes.

Can I say something about the normality of the $p_i$-Sylow subgroups? Can I say for instance that $P_n$ is normal?

I know then that if $P_1$ is a $p_1$- Sylow subgroup than $G$ has a $p_1$ normal complement K.

Is it true that $K$ is cyclic? How can I show that?

Thanks for the help!

• No, $K$ need not be cyclic (this is easy to check by considering some examples). The Sylow-$p_n$ subgroup will be normal by induction, and the group will be an iterated semidirect product of the Sylows in descending order. – Tobias Kildetoft Mar 7 '16 at 8:56
• @TobiasKildetoft Thank you! I don't understand very well how can I prove the normality of $P_n$ by induction – Gggl. Mar 7 '16 at 9:11
• You need to use (or at least this makes it a lot easier) that normal Hall subgroups are characteristic, so you can use transitivity. – Tobias Kildetoft Mar 7 '16 at 9:30
• @TobiasKildetoft Ah ok..thank you so much for the help! – Gggl. Mar 7 '16 at 9:35
• By the way, these groups are sometimes call $Z$-groups, though I am unsure how common the term is. – Tobias Kildetoft Mar 7 '16 at 9:37

In addition to the remarks of Tobias (with $G=S_3$ you can refute your statement), one can prove that if a group has cyclic Sylow subgroups, it must be solvable (of derived length $\leq 2$). For a proof see for example M.I. Isaacs, Finite Group Theory, Corollary 5.15
• $S_3$ does not refute the normal complement being cyclic. The group is even supersolvable (though this is rarely as relevant), and thus an $M$-group (and also an IPR-group though probably nobody but I care about those). – Tobias Kildetoft Mar 7 '16 at 9:44
• Ahh, I don't seem to see any claims of them all being normal, just a question about what can be said about their normality and whether the $p_n$ one is normal. – Tobias Kildetoft Mar 7 '16 at 9:48
• Ah right, I read $G$ being cyclic, where it says $K$. – Nicky Hekster Mar 7 '16 at 9:57