# If $H$ is maximal subgroup of $A_{34}$ of index $34$, then $H\cong A_{33}$?

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.

• The number of Sylow 2-subgroups of any group is odd. It cannot be $34$. – Mark Sapir Oct 3 '20 at 0:54
• @JCAA, Thank you, I have did a correction. – Yi Wang Oct 3 '20 at 0:58
• What exactly are you asking? – Derek Holt Oct 3 '20 at 7:56
• @Derek Holt, I have made some additions, I hope you can know what I want to ask. – Yi Wang Oct 4 '20 at 0:33
• 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. – Derek Holt Oct 4 '20 at 8:13

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 $$A_{33}$$.

• Wouldn't the homomorphism go to $S_{33}$? And, where is maximality used? – user403337 Oct 3 '20 at 1:06
• No, the index is $34$. So the homomorphism will map into $S_{34}$. – Mark Sapir Oct 3 '20 at 1:08
• Sorry about that. – user403337 Oct 3 '20 at 1:08
• 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. – Mark Sapir Oct 3 '20 at 3:10
• For the statement about subgroups of index $n$ see this question: math.stackexchange.com/questions/1230037/… – Mark Sapir Oct 4 '20 at 3:31

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)

• Thank you very much! – Yi Wang Oct 4 '20 at 3:32