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I've this exercise to resolve : prove that if $G$ is a finite group and all its Sylow subgroups are cyclic then G is supersoluble. My solution follows: is it correct? Thanks to everyone for the help!

Let's proceed for induction on $|G|$. Let $p_1$$<$$p_2$$...$$<$$p_n$ the distinct primes that factorise $|G|$ and let $P_i$ be the corresponding p-Sylow's subgroups. Since $p_1$ is the smallest prime dividing $|G|$ then exists $K$$\vartriangleleft$ $G$ such that $G$$=$$K$$\rtimes$$P_1$ (hence $K$$=$$P_2$$P_3$$...$$P_n$). Now applying to $K$ the inductive hypothesis, $1$$\lt$$K_1$$\lt$$...$$\lt$$K_s$$\lt$$K$ is a normal series whose factors have prime order. On the other hand $G/K$$\simeq$$P_1$ is a finite abelian p-group so I can consider each subgroup $H_i/K$$\leq$$G/K$ with $|H_i/K|$$=$$p^i_1$ for $i$$=$$1,...,n-1$ and $|P_1|$$=$$p_1^n$. In conclusion $1$$\lt$$K_1$$\lt$$...$$\lt$$K_s$$\lt$$K$$\lt$$H_1$$...$$\lt$$H_{n-1}$$\lt$$G$ is a normal series for $G$ whose factors have prime order.

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Sibilla
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  • Yes, this all looks fine. The hard part of this is really to recall that having a cyclic $p$-Sylow for the smallest prime dividing the order leads to a normal $p$-complement. – Tobias Kildetoft Jan 23 '16 at 20:47
  • @TobiasKildetoft Yes, this exercise is proposed on my book as application of the theorem you've mentioned. Thanks for the help! – Sibilla Jan 23 '16 at 20:55
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    I don't think that you have proved that the $K_i$ are normal in $G$. You have only proved that they are normal in $K$. – Derek Holt Jan 23 '16 at 20:56
  • @DerekHolt You've right! The fact that $K$ as normal p-complement in $G$ is a characteristic subgroup is useful to complete the proof? Thanks for the help! – Sibilla Jan 23 '16 at 21:13
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    Yes you could show that $P_n$ is normal and hence characteristic in $G$, and all subgroups of a cyclic group are characteristic, so the subgroups of a normal series of $P_n$ are normal in $G$. Then apply induction to $G/P_n$. – Derek Holt Jan 23 '16 at 21:38
  • Alternatively, note that each normal complement you find is a normal Hall subgroup and thus characteristic. – Tobias Kildetoft Jan 24 '16 at 10:09
  • @Derek Thank you very much! You've been very helpful! I should have applied the inductive hypotesis to the factor $G/P_n$! – Sibilla Jan 24 '16 at 12:16

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