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Problem: If $G$ is a finite group whose Sylow $p$-subgroups are all cyclic then $G$ has normal subgroup $N$ and such that $G/N$ and $N$ are both cyclic.

Whenever I need to find normal subgroup, I always try to find one homomorphim $\phi$ to show $\text{ker}\:\phi$ is non-trivial for a homomorphism $\phi$ on $G$.

But for this problem, I couldn't come up with any trivial one. I guess some tricky observation is needed. Any hint will be appreciated.

Nicky Hekster
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N00BMaster
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  • Hint: There is only one natural choice of $N$, namely the Fitting subgroup $F(G)$ of $G$ (as the largest normal nilpotent subgroup of $G$, in this case it is the largest normal cyclic subgroup of $G$). – user10354138 Oct 22 '23 at 21:34
  • We haven't introduced Fitting subgroup. But as you said largest normal cyclic subgroup, then we need to take all the sylow p subgroups (all conjugacy class) as $N$. Then $N$ is normal (what about cyclic?) Can you give some more hint on $G/N$? @user10354138 and thanks for your response – N00BMaster Oct 23 '23 at 05:50
  • Hints: (1) $G$ is solvable. (2) a finite nilpotent group is the direct product of its Sylows. (3) the Fitting subgroup of a solvable group is self-centralising. (4) the automorphism group of a cyclic group is cyclic. – user10354138 Oct 23 '23 at 06:14
  • I'm concerned that I might not be able to complete this using your hints . I don't see why we need to use solvable and fitting subgroups here, especially since I'm just starting this problem after finishing the Sylow theorems, @user10354138 – N00BMaster Oct 23 '23 at 06:26
  • @user10354138 It is not true that the automorphism groups of all cyclic groups are cyclic. For example ${\rm Aut}(C_8) \cong C_2 \times C_2$. – Derek Holt Oct 23 '23 at 06:32
  • Oops, I meant automorphism group is abelian (and so $G/F(G)$ cyclic when applied to this case). – user10354138 Oct 23 '23 at 06:37
  • I will try to study the mentioned topics first. Thank you all. Can you suggest something on this question? @DerekHolt – N00BMaster Oct 24 '23 at 08:32

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This is a result of Hölder, Burnside, and Zassenhaus. See 10.1.10 in Robinson "A Course in the Theory of Groups". The proof uses various tools such as Burnside's transfer theorem that are rather more sophisticated than just the Sylow theorems, so this "exercise" does not seem appropriate to your level of knowledge.

As a special case, if $G$ is simple and nonabelian, the exercise implies that $G$ has a non-cyclic Sylow subgroup. How are you supposed to prove this using just the Sylow theorems?

Sean Eberhard
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  • 10.1.10 in Robinson provides much more than asked, namely the presentation of a group with only cyclic Sylow subgroups. – Nicky Hekster Oct 23 '23 at 14:35
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    Hall's book (Theorem 9.4.3) uses a counting argument only (also uses: showing $G$ solvable, and use $G/G', G'/G''$ cyclic implies $G''=1$) and not Burnside transfer, but since OP doesn't even want to get near solvable groups and derived series it is probably also not appropriate. – user10354138 Oct 23 '23 at 16:54
  • I think I need to wait until the lecturer complete the mentioned topics, @user10354138, NickyHekster Thank you all. Can you suggest something on this question? – N00BMaster Oct 24 '23 at 08:29