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Is there any example of finite group $G$ with the following properties?

1) There is prime divisor $p$ of order $G$ such that the number of cyclic subgroup of order $p$ is $p+1$.

2) The order of Sylow $p$-subgroup of $G$ is $p^{2}$.

3) $G$ is not $p$-group.

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4  
I believe that the Klein 4-group qualifies. In this case, $p = 2$, and we have 3 subgroups of order 2, and $G$ is its own Sylow 2-subgroup. –  David Wheeler Jul 22 '12 at 7:40
1  
To expand David Wheeler's example, you can take $(\mathbb Z/p\mathbb Z)^2$ for any prime number $p$. –  Marc van Leeuwen Jul 22 '12 at 8:03
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@Niki: If that is a third requirement, you should edit it into your question. –  Mikko Korhonen Jul 22 '12 at 8:26
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... And now $(\mathbb{Z}/p\mathbb{Z})^2 \times H$ is a solution, whenever $p$ is prime and $H$ is nontrivial of order coprime to $p$. Unless there are still more conditions you haven't mentioned? –  Chris Eagle Jul 22 '12 at 9:02

3 Answers 3

up vote 2 down vote accepted

Let $G$ be a group with all the properties stated in your question, and let P be a Sylow $p$-subgroup. If $P$ is not cyclic, then $P$ must be elementary Abelian, hence already contains $p+1$ subgroups of order $p.$ Then $P$ must be normal, otherwise a conjugate $Q$ of $P$ different from $P$ will contain another subgroup of order $p$ not contained in $P.$ On the other had, any group with a (proper) normal Sylow $p$-subgroup which is elementary Abelian of order $p^{2}$ will satisfy your conditions.

Suppose then that $P$ is cyclic of order $p^{2}.$ I will derive a contradiction. Notic that the conjugation action of $G$ on its subgroups of order $p$ gives a homomorphism from $G$ to the symmetric group $S_{p+1}.$ Since $S_{p+1}$ does not contain elements of order $p^{2},$ $P$ must intersect the kernel of this homomorphism non-trivially. Let $Q$ be the unique subgroup of $P$ of order $p.$ Then $Q$ normalizes each subgroup of $G$ of order $p.$ If $R$ is one of these, but $R \neq Q,$ then $RQ$ is a group of order $p^{2}$ containing more than one subgroup of order $p,$ so must be Abelian of order $p^{2},$ so must be a Sylow $p$-subgroup of $G$, hence conjugate to $P,$ a contradiction, as $P$ is not cyclic. Hence $Q$ must be the only subgroup of $G$ of order $p,$ again a contradiction.

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DouglasS.Stones and DonAntonio: Thank you very much, they were most helpful! –  N K Jul 22 '12 at 11:15

Your question's been answered in general already, yet if you want a specific "easy" example take $\,G=A_4\,\,,\,\,p=2\,$ .

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Here's a snippet of GAP code that can search for the groups you request (it uses the SONATA package):

RequirePackage("sonata");

for p in Filtered([2..5],IsPrimeInt) do
  for r in Filtered([2..5],i->(not IsPrimePowerInt(i)) or i mod p<>0) do
    n:=p^2*r;
    for G in AllSmallGroups(n) do
      S:=Filtered(Subgroups(G),H->Size(H)=p);
      if(Size(S)=p+1 and Size(SylowSubgroup(G,p))=p^2) then
        Print("Group: ",StructureDescription(G)," of order ",Size(G)," has p+1 subgroups of order p and Sylow p-subgroup of order p^2, where p=",p,"\n");
      fi;
    od;
  od;
od;

This code implicitly uses the property that (a) all groups of prime order a cyclic, and (b) Lagrange's Theorem (so $p^2$ must divide the order of the group).

Here's the output:

Group: A4 of order 12 has p+1 subgroups of order p and Sylow p-subgroup of order p^2, where p=2
Group: C6 x C2 of order 12 has p+1 subgroups of order p and Sylow p-subgroup of order p^2, where p=2
Group: C10 x C2 of order 20 has p+1 subgroups of order p and Sylow p-subgroup of order p^2, where p=2
Group: C3 x S3 of order 18 has p+1 subgroups of order p and Sylow p-subgroup of order p^2, where p=3
Group: (C3 x C3) : C2 of order 18 has p+1 subgroups of order p and Sylow p-subgroup of order p^2, where p=3
Group: C6 x C3 of order 18 has p+1 subgroups of order p and Sylow p-subgroup of order p^2, where p=3
Group: C3 x (C3 : C4) of order 36 has p+1 subgroups of order p and Sylow p-subgroup of order p^2, where p=3
Group: (C3 x C3) : C4 of order 36 has p+1 subgroups of order p and Sylow p-subgroup of order p^2, where p=3
Group: C12 x C3 of order 36 has p+1 subgroups of order p and Sylow p-subgroup of order p^2, where p=3
Group: (C3 x C3) : C4 of order 36 has p+1 subgroups of order p and Sylow p-subgroup of order p^2, where p=3
Group: S3 x S3 of order 36 has p+1 subgroups of order p and Sylow p-subgroup of order p^2, where p=3
Group: C6 x S3 of order 36 has p+1 subgroups of order p and Sylow p-subgroup of order p^2, where p=3
Group: C2 x ((C3 x C3) : C2) of order 36 has p+1 subgroups of order p and Sylow p-subgroup of order p^2, where p=3
Group: C6 x C6 of order 36 has p+1 subgroups of order p and Sylow p-subgroup of order p^2, where p=3
Group: C15 x C3 of order 45 has p+1 subgroups of order p and Sylow p-subgroup of order p^2, where p=3
[[snipped]]

(Here ":" denotes a semidirect product.)

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