# Question on the unsolvability a group

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.

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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 $2^{22} 3^4 5^2$.

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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:

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

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

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