Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $G$ be a finite group. Let $\pi(G)=\{2,3,5\}$ be the set of prime divisors of its order. If 6 divide the number of Sylow 5-subgroups of G and 10 divide the number of Sylow $3$-subgroups of $G$, then whether the group $G$ group with those properties is unsolvable?

In particular if the number of Sylow $5$-subgroups of $G$ is 6 or the number of Sylow $3$-subgroups of $G$ is 10, then by the Hall's theorem $G$ is unsolvable group. For example if the number of Sylow $5$-subgroups of $G$ is 6 and $G$ is solvable, then $2\equiv 1$ (mod $5$), a contradiction.

Hall's theorem: Let $G$ be a finite soluble group and $|G|=m.n$, where $m=p_{1}^{\alpha _{1}}...p_{r}^{\alpha _{r}}$, $(m,n)=1$. Let $\pi =\{p_{1},...,p_{r}\}$ and $ h_{m}$ be the number of $\pi -$Hall subgroups of $G$. Then $ h_{m}=q_{1}^{\beta _{1}}...q_{s}^{\beta _{s}}$, satisfies the following conditions for all $i\in \{1,2,...,s\}$:

1) $q_{i}^{\beta_{i}} \equiv 1$ (mod $p_{j}$), for some $p_{j}$.

2) The order of some chief factor of $G$ is divisible by $ q_{i}^{\beta_{i}}$.

Thank you so much.

share|improve this question

2 Answers 2

up vote 2 down vote accepted

No. There is a solvable group of order $2^{22} 3^5 5^3$ with $n_2=1$, $n_3=2^{18} 5^2$, and $n_5 = 2^{20} 3^4$.$\newcommand{\GF}{\operatorname{GF}}\newcommand{\AGL}{\operatorname{AGL}}$

$$G = \left(\left(3\ltimes\GF(5^2)\right) \ltimes \left(\GF(2^4)^3\right)\right) ~ \times ~ \left(\left(5\ltimes\GF(3^4)\right) \ltimes \left(\GF(2^2)^5\right)\right)$$

I'll describe the pieces:

$H = 3\ltimes\GF(5^2) \leq \AGL(1,5^2)$ is the collection of affine maps $x\mapsto ax+b$ where $a,x,b$ are elements of the finite field $\GF(5^2)$ of order 25 and $a^3=1$.

$H$ has an irreducible $\GF(2^4)$ module $V$ of dimension 3 formed by inducing a non-principal one-dimensional module from $\GF(5^2) \leq H$.

$H \ltimes V =\left(3\ltimes\GF(5^2)\right) \ltimes \left(\GF(2^4)^3\right) \leq \AGL(3,2^4)$ consists of all maps $x\mapsto ax+b$ where $a \in H$ and $x,b \in V$.

$K = 5 \ltimes \GF(3^4) \leq \AGL(1,3^4)$ is the collection of affine maps $x\mapsto ax+b$ where $a,x,b$ are elements of the finite field $\GF(3^4)$ of order 81 and $a^5=1$.

$K$ has an irreducible $\GF(2^2)$ module $W$ of dimension 5 formed by inducing a non-principal one-dimensional module from $\GF(3^4) \leq K$.

$K \ltimes W = \left(5\ltimes\GF(3^4)\right) \ltimes \left(\GF(2^2)^5\right) \leq \AGL(5,2^2)$ consists of all maps $x\mapsto ax+b$ where $a \in K$ and $x,b \in W$.

$G = \left( H \ltimes V \right) \times \left( K \ltimes W \right)$ is the direct product of these two groups.

$\begin{array}{c|cccc} & o & n_2 & n_3 & n_5 \\ \hline H & 2^{~0} 3^1 5^2 & 3^0 5^0 & 2^{~0} 5^2 & 2^{~0} 3^0 \\ H \ltimes V & 2^{12} 3^1 5^2 & 3^0 5^0 & 2^{~8} 5^2 & 2^{12} 3^0 \\ K & 2^{~0} 3^4 5^1 & 3^0 5^0 & 2^{~0} 5^0 & 2^{~0} 3^4 \\ K \ltimes W & 2^{10} 3^4 5^1 & 3^0 5^0 & 2^{10} 5^0 & 2^{~8} 3^4 \\ G & 2^{22} 3^5 5^3 & 3^0 5^0 & 2^{18} 5^2 & 2^{20} 3^4 \\ \end{array}$

I think this is approximately minimal order. I think I misunderstand Hall's theorem if there is any example of order less than $2^{22} 3^4 5^2$.

share|improve this answer
    
Thanks, it is very nice example. Do you think when $n_{2}\neq 1$ there is any example? –  N K Jun 18 '12 at 4:14
    
$G \times D$ where $D$ is dihedral of order 30. –  Jack Schmidt Jun 18 '12 at 12:15
    
Thank you so much. –  N K Jun 18 '12 at 13:29

Yup, I misunderstood Hall's theorem a little. There was no need to fix the 5-part of $n_3$ before doing the 2-part, and no need for the reps to be irreducible.

Here is a much smaller example constructed using only the obvious idea from Hall.$\newcommand{\GF}{\operatorname{GF}}\newcommand{\AGL}{\operatorname{AGL}}$

Let $G = H \times K \times L$ where $H,K$ are as before, and $L$ is similar.

  • $H = 3 \ltimes \GF(5^2) \leq \AGL(1,5^2)$
  • $K = 5 \ltimes \GF(3^4) \leq \AGL(1,3^4)$
  • $L = 15 \ltimes \GF(2^4) = \AGL(1,2^4)$

$\begin{array}{c|cccc} & o & n_2 & n_3 & n_5 \\ \hline H & 2^0 3^1 5^2 & 3^0 5^0 & 2^0 5^2 & 2^0 3^0 \\ K & 2^0 3^4 5^1 & 3^0 5^0 & 2^0 5^0 & 2^0 3^4 \\ L & 2^4 3^1 5^1 & 3^0 5^0 & 2^4 5^0 & 2^4 3^0 \\ G & 2^4 3^6 5^4 & 3^0 5^0 & 2^4 5^2 & 2^4 3^4 \\ \end{array}$

One can also take the full AGLs in order to get $n_2$ divisible by 15:

  • $H_0 = 24 \ltimes \GF(5^2) = \AGL(1,5^2)$
  • $K_0 = 80 \ltimes \GF(3^4) = \AGL(1,3^4)$
  • $L_0 = 15 \ltimes \GF(2^4) = \AGL(1,2^4)$
  • $G_0 = H_0 \times K_0 \times L_0$

$\begin{array}{c|cccc} & o & n_2 & n_3 & n_5 \\ \hline H_0 & 2^3 3^1 5^2 & 3^0 5^2 & 2^0 5^2 & 2^0 3^0 \\ K_0 & 2^4 3^4 5^1 & 3^4 5^0 & 2^0 5^0 & 2^0 3^4 \\ L_0 & 2^4 3^1 5^1 & 3^0 5^0 & 2^4 5^0 & 2^4 3^0 \\ G_0&2^{11}3^6 5^4 & 3^4 5^2 & 2^4 5^2 & 2^4 3^4 \\ \end{array}$

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.