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Theorem:Let $P\in Syl_p(G)$ for a finite group $G$. Then $G$ has a normal $p-$ complement if and only if $P$ controls its own fusion.

I wonder if similar argument is true for Hall subgroups (in general or in solvable groups)? If yes, is there any reference ?

I had asked it there but I did not take any response.

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The result is true for Hall subgroups in solvable groups, but not in general. I don't have the references to hand, but it's a Theorem of Brauer, or maybe E.C. Dade, or maybe Suzuki, that if $G$ has a Hall $\pi$-subgroup $H$ which controls the fusion of its elements AND every Brauer elementary $\pi$-subgroup of $G$ is conjugate to a subgroup of $H$, then $G$ has a normal $\pi$-complement ( the conditions are also necessary). By Hall's Theorems, the condition on Brauer elementary Subgroups is automatically satisfied if $G$ is solvable ( probably $\pi$-solvable will do).

The proof is as follows. Take an irreducible character $\chi$ of $H$: extend it to a well-defined class-function $\chi^{\ast}$ of $G$ as follows: let $g_{\pi}$ be the $\pi$-part of $g$, and set $\chi^{\ast}(g) = \chi(h),$ where $h$ is any element of $H$ which is conjugate to $g_{\pi}.$ Then by Brauer's characterization of characters, $\chi^{\ast}$ is a generalized character of $G$, using the second assumption. The second condition also ensures (via a counting argument) that $\chi^{\ast}$ is irreducible. Then the normal complement is $\bigcap_{\chi \in {\rm Irr}(H)} {\rm ker}(\chi^{\ast}).$ I think there is a proof in W. Feit's book "Characters of Finite Groups" if more detail is required.

However, if $p$ is a prime greater than $3,$ note that the symmetric group $S_{p}$ has a Hall $p^{\prime}$-subgroup ( namely $S_{p-1}$) which controls the fusion of its elements ( for two elements are conjugate if and only if they have the same cycle type), but there is evidently no normal complement. It is easy to check that there are $p$-regular elements of $S_{p}$ which are not conjugate to any element of $S_{p-1}$ ( choose a product of disjoint $2$- and $3$-cycles which moves all points, for example), so the condition on Brauer elementary $p^{\prime}$-subgroups is not satisfied.

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  • $\begingroup$ Dear Geoff Robinson thank you for your information and counter example but I need reference or a proof. $\endgroup$
    – mesel
    Commented Oct 19, 2016 at 17:37
  • $\begingroup$ You have the proof in the countererexample. I will include the proof in my answer now for the other part. $\endgroup$
    – Geoff Robinson
    Commented Oct 19, 2016 at 17:40
  • $\begingroup$ By the way, I have interpreted "controls its own fusion" as "controls the $G$-fusion of its elements". There are other possible interpretations. $\endgroup$
    – Geoff Robinson
    Commented Oct 19, 2016 at 23:13
  • $\begingroup$ Yes by saying controls its own fusion I mean controls the G-fusion of its elements. Thank you very much. I had thought that it has a proof with Transfer theory and may be induction on the order of G for solvable group. $\endgroup$
    – mesel
    Commented Oct 20, 2016 at 9:59
  • $\begingroup$ In fact, the $G = S_{5},H = S_{4}$ example shows what can go wrong (although $G$ itself is not solvable). In that example, $G^{\prime } = A_{5}$ and $H^{\prime} = A_{4}.$ It is still true that $H^{\prime}$ is a Hall $\{2,3\}$-subgroup of $G^{\prime},$ but $H^{\prime}$ no longer controls the fusion of its elements in $G^{\prime},$ since an element of order $3$ in $H^{\prime}$ is conjugate to its inverse in $G^{\prime},$ but not in $H^{\prime}.$ $\endgroup$
    – Geoff Robinson
    Commented Oct 20, 2016 at 16:50

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