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Let $A_{34}$ be the alternating group of degree 34, $G_3$ be a Sylow $3$-subgroup of $A_{34}$ and $H=N_{A_{34}}(G_3)$. If $H$ is maximal subgroup of $A_{34}$ of index $34$, then $H\cong A_{33}$?

I want to prove that $A_{34}$ does not have exactly $34$ Sylow $3$-subgroups. If $A_{34}$ has $34$ Sylow $3$-subgroups, then $|G:N_G(G_3)|=34=2.17$, where $G=A_{34}$ and $G_3$ is a Sylow $3$-subgroup of $A_{34}$. For some reasons, we can prove that $H=N_G(G_3)$ is a maximal subgroup of $G=A_{34}$ of index $34$. Now if we can prove that $H=N_G(G_3)\cong A_{33}$, then we get a contradiction as $A_{33}$ is a simple group, hence $A_{34}$ does not have $34$ Sylow $3$-subgroups.

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    $\begingroup$ The number of Sylow 2-subgroups of any group is odd. It cannot be $34$. $\endgroup$ – Mark Sapir Oct 3 '20 at 0:54
  • $\begingroup$ @JCAA, Thank you, I have did a correction. $\endgroup$ – Yi Wang Oct 3 '20 at 0:58
  • $\begingroup$ What exactly are you asking? $\endgroup$ – Derek Holt Oct 3 '20 at 7:56
  • $\begingroup$ @Derek Holt, I have made some additions, I hope you can know what I want to ask. $\endgroup$ – Yi Wang Oct 4 '20 at 0:33
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    $\begingroup$ No it's not. Are you asking "How do we prove that if $H$ is a maximal subgroup of $A_{34}$ with index $34$ then $H \cong A_{33}$?", or are asking for help with proving that $A_{34}$ does not have (exactly?) $34$ Sylow $3$-subgroups? They are two different questions. $\endgroup$ – Derek Holt Oct 4 '20 at 8:13
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Yes. Consider the Schreier action of $A_{34}$ on $A_{34}/H$. It gives a homomorphism from $A_{34}$ to $S_{34}$. Since $A_{34}$ is simple the image is $A_{34}$ with the natural action on a $34$-element set. Now $H$ is the stabilizer of a point of that action. So it is isomorphic to $A_{33}$.

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  • $\begingroup$ Wouldn't the homomorphism go to $S_{33}$? And, where is maximality used? $\endgroup$ – user403337 Oct 3 '20 at 1:06
  • $\begingroup$ No, the index is $34$. So the homomorphism will map into $S_{34}$. $\endgroup$ – Mark Sapir Oct 3 '20 at 1:08
  • $\begingroup$ Sorry about that. $\endgroup$ – user403337 Oct 3 '20 at 1:08
  • $\begingroup$ If the subgroup of index $34$ is not maximal then there is a maximal sbgroup of index $17$ or $2$. Both cases are not possible. $\endgroup$ – Mark Sapir Oct 3 '20 at 3:10
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    $\begingroup$ For the statement about subgroups of index $n$ see this question: math.stackexchange.com/questions/1230037/… $\endgroup$ – Mark Sapir Oct 4 '20 at 3:31
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We can test this with the help of Magma:

>A34:= AlternatingGroup(34);
>G3:=Sylow(A34,3);
>G3;
>H:=Normalizer(A34, G3);
>H;
>IsMaximal(A34, H);
>Subgroups(A34: Al := "Maximal");

The function Sylow is explained as follows:

Sylow(G, p) : GrpPerm, RngIntElt -> GrpPerm

Given a group G and a prime p, construct a Sylow p-subgroup of G. The algorithm used is that of Cannon, Cox and Holt [CCH97].

We get the following results for Normalizer(A34, G3):

Permutation group acting on a set of cardinality 34
Order = 459165024 = 2^5 * 3^15
    (1, 3, 2)(4, 6, 5)(10, 11, 12)(22, 23, 24)
    (1, 2, 3)
    (1, 2, 3)(4, 6, 5)(10, 11, 12)(19, 21, 20)(25, 27, 26)(28, 29, 30)(31, 33,
        32)
    (1, 2, 3)(19, 20, 21)(28, 30, 29)(31, 33, 32)
    (1, 3, 2)(10, 12, 11)(13, 15, 14)
    (7, 16, 25)(8, 17, 26)(9, 18, 27)(10, 19, 28)(11, 20, 29)(12, 21, 30)(13,
        22, 31)(14, 23, 32)(15, 24, 33)
    (2, 3)(10, 13)(11, 14)(12, 15)(19, 22)(20, 23)(21, 24)(28, 32)(29, 33)(30,
        31)
    (1, 2, 3)(13, 14, 15)(16, 23, 19, 17, 24, 20, 18, 22, 21)(25, 31, 30, 27,
        33, 29, 26, 32, 28)
    (4, 5, 6)
    (1, 3, 2)(4, 5, 6)(10, 12, 11)(13, 14, 15)(19, 21, 20)(22, 23, 24)(25, 29,
        32, 26, 30, 33, 27, 28, 31)
    (1, 3, 2)(10, 11, 12)(13, 15, 14)(16, 18, 17)(22, 24, 23)(25, 30, 31)(26,
        28, 32)(27, 29, 33)
    (7, 8, 9)
    (10, 11, 12)(13, 14, 15)(19, 20, 21)(31, 33, 32)
    (1, 2, 3)(10, 12, 11)(13, 15, 14)(19, 20, 21)(22, 24, 23)
    (13, 15, 14)
    (1, 2, 3)(10, 12, 11)(13, 15, 14)
    (1, 3, 2)(10, 12, 11)(13, 15, 14)(19, 20, 21)(22, 24, 23)
    (7, 10, 13)(8, 11, 14)(9, 12, 15)
    (2, 3)(8, 9)(11, 12)(14, 15)(17, 18)(20, 21)(23, 24)(25, 27)(29, 30)(31, 33)
    (1, 2, 3)(4, 6, 5)(10, 11, 12)(22, 23, 24)
    (8, 9)(11, 12)(14, 15)(16, 26)(17, 25)(18, 27)(19, 28)(20, 30)(21, 29)(22,
        32)(23, 31)(24, 33)
    (1, 2, 3)(10, 11, 12)(13, 15, 14)(16, 18, 17)(22, 24, 23)(25, 30, 31)(26,
        28, 32)(27, 29, 33)
    (2, 3)(5, 6)
    (1, 3, 2)(4, 6, 5)(10, 11, 12)(19, 21, 20)(25, 27, 26)(28, 29, 30)(31, 33,
        32)
    (1, 4)(2, 5, 3, 6)
    (1, 3, 2)(19, 20, 21)(28, 30, 29)(31, 33, 32)
    (1, 2, 3)(4, 5, 6)(10, 12, 11)(13, 14, 15)(19, 21, 20)(22, 23, 24)(25, 29,
        32, 26, 30, 33, 27, 28, 31)
    (1, 3, 2)(13, 14, 15)(16, 23, 19, 17, 24, 20, 18, 22, 21)(25, 31, 30, 27,
        33, 29, 26, 32, 28)
$\endgroup$
1
  • $\begingroup$ Thank you very much! $\endgroup$ – Yi Wang Oct 4 '20 at 3:32

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