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September 10th, 2017, 12:15 PM  #1 
Senior Member Joined: Jan 2015 From: usa Posts: 103 Thanks: 1  action of Frobenius group
Let $p$ be a prime number. Show that the Frobenius group $F_{p(p1)}$ acts on the set $\mathbb{F}_p$ by: $$\begin{pmatrix}a&b \\\ 0&1\end{pmatrix}x:=ax+b$$ 
September 10th, 2017, 01:11 PM  #2 
Member Joined: May 2017 From: Russia Posts: 34 Thanks: 5 
$\displaystyle \left(\begin{pmatrix}a_1&b_1 \\\ 0&1\end{pmatrix}\begin{pmatrix}a_2&b_2 \\\ 0&1\end{pmatrix}\right)x = \begin{pmatrix}a_1a_2&a_1b_2+b_1 \\\ 0&1\end{pmatrix}x= a_1a_2x+a_1b_2+b_1.$ $\displaystyle \left(\begin{pmatrix}a_1&b_1 \\\ 0&1\end{pmatrix}\begin{pmatrix}a_2&b_2 \\\ 0&1\end{pmatrix}\right)x=a_1(a_2x+b_2)+b_1.$ 
September 10th, 2017, 01:49 PM  #3 
Senior Member Joined: Jan 2015 From: usa Posts: 103 Thanks: 1 
And how does that answer the quetion? Can you explain to me more? Thanks in advance.
Last edited by skipjack; September 11th, 2017 at 02:56 PM. 
September 10th, 2017, 03:36 PM  #4 
Member Joined: May 2017 From: Russia Posts: 34 Thanks: 5 
The set of polynomials $\displaystyle ax+b$, where $\displaystyle a,b\in\mathbb{F}_p$ and $\displaystyle a\neq 0$, is a group (the operation is composition $\displaystyle \circ$). The group acts on $\displaystyle \mathbb{F}_p$: $\displaystyle ax+b:\ t\mapsto at+b, \ \ \ t\in \mathbb{F}_p$. And $\displaystyle (ax+b)(\mathbb{F}_p)=\mathbb{F}_p$. It permutes the elements of $\displaystyle \mathbb{F}_p$. One has isomorphism $\displaystyle \varphi$ from the group of matrices $\displaystyle \begin{pmatrix}a&b \\\ 0&1\end{pmatrix}$ to the group of $\displaystyle ax+b$: $\displaystyle \varphi:\ \begin{pmatrix}a&b \\\ 0&1\end{pmatrix} \mapsto ax+b$. $\displaystyle \varphi\left(\begin{pmatrix}a_1&b_1 \\\ 0&1\end{pmatrix}\begin{pmatrix}a_2&b_2 \\\ 0&1\end{pmatrix}\right)=a_1a_2x+a_1b_2+b_1=a_1(a_2 x+b_2)+b_1= (a_1x+b_1)\circ(a_2x+b_2)= $ $\displaystyle =\varphi\left(\begin{pmatrix}a_1&b_1 \\\ 0&1\end{pmatrix}\right) \circ\varphi\left(\begin{pmatrix}a_2&b_2 \\\ 0&1\end{pmatrix}\right). $ 
September 10th, 2017, 08:09 PM  #5 
Senior Member Joined: Jan 2015 From: usa Posts: 103 Thanks: 1 
Thanks for the explanation. and is it possible to show that this action is sharply 2transitive i.e. if $x_1,x_2\in\mathbb{F}_P$ are distinct and $x_1,x_2\in\mathbb{F}_P$ are distinct then there is exactly one $g\in F_{p(p1)}$ such that $gx_{1}=y_1$ and $gx_{2}=y_2$ ?
Last edited by skipjack; September 11th, 2017 at 02:56 PM. 
September 11th, 2017, 05:10 AM  #6 
Member Joined: May 2017 From: Russia Posts: 34 Thanks: 5 
Let $\displaystyle g=\begin{pmatrix}a&b \\\ 0&1\end{pmatrix}$. If $\displaystyle gx_{1}=y_1$ and $\displaystyle gx_2=y_2$ then a,b have to be the solution of the system $\displaystyle \begin{cases} x_1\cdot a + b = y_1,\\ x_2\cdot a +b = y_2. \end{cases}$ This system has unique solution. And $\displaystyle a\neq 0$, otherwise we would have $\displaystyle y_1=y_2$. Let $\displaystyle x_3\in \mathbb{F}_p$ and $\displaystyle x_3\neq x_1,x_2$. $\displaystyle g$ necessarily sends x_3 to $\displaystyle ax_3+b$, not to an arbitrary element one could choose. 
September 11th, 2017, 12:00 PM  #7 
Senior Member Joined: Jan 2015 From: usa Posts: 103 Thanks: 1 
In fact, I have a quiz next Friday and I tried to identify the center $Z(F_{p(p1)})$ and $F'_{p(p1)}=[F_{p(p1)},F_{p(p1)}]$ with familiar groups but I didn't manage to do that. Can you please help me? I will be grateful if you could help me.
Last edited by skipjack; September 11th, 2017 at 02:53 PM. 
September 12th, 2017, 03:09 AM  #8 
Member Joined: May 2017 From: Russia Posts: 34 Thanks: 5 
Let $\displaystyle H=\left\{ \begin{pmatrix}1&b \\\ 0&1\end{pmatrix}\ \bigg \ b \in \mathbb{F}_p \right\}$. $\displaystyle H$ is a normal subgroup in $\displaystyle F_{p(p1)}$. The order of $\displaystyle H$ is equal to $\displaystyle p$. Find the commutator of $\displaystyle \begin{pmatrix} a&b \\\ 0&1\end{pmatrix},\begin{pmatrix}c&d \\\ 0&1\end{pmatrix}\in F_{p(p1)}$: $\displaystyle \begin{pmatrix} a&b \\\ 0&1\end{pmatrix}\begin{pmatrix}c&d \\\ 0&1\end{pmatrix}\begin{pmatrix} \frac{1}{a} & \frac{a}{b} \\\ 0&1\end{pmatrix}\begin{pmatrix}\frac{1}{c}& \frac{d}{c} \\\ 0&1\end{pmatrix} = \begin{pmatrix} ac& ad+b \\\ 0&1\end{pmatrix}\begin{pmatrix}\frac{1}{ac}& \frac{d}{ac}\frac{b}{a} \\\ 0&1\end{pmatrix}=\begin{pmatrix} 1& \ldots \\\ 0&1\end{pmatrix}$. So, $\displaystyle H$ contains every commutator of $\displaystyle F_{p(p1)}$. If a group is not commutative, its commutator subgroup is not trivial. The commutator subgroup $\displaystyle [F_{p(p1)},F_{p(p1)}]$ is a subgroup of $\displaystyle H$. The order of $\displaystyle [F_{p(p1)},F_{p(p1)}]$ is a divisor of the order of $\displaystyle H$ which is prime. So, if $\displaystyle p\gt 2$ the order of the commutator group is equal to $\displaystyle p$, and $\displaystyle [F_{p(p1)},F_{p(p1)}]=H$. If $\displaystyle p=2$ then $\displaystyle F_{p(p1)}$ is commutative and its commutator subgroup is trivial. Last edited by ABVictor; September 12th, 2017 at 03:14 AM. 
September 12th, 2017, 07:01 AM  #9 
Senior Member Joined: Jan 2015 From: usa Posts: 103 Thanks: 1 
and can us identify $F_{p(p1)}/F'_{p(p1)}$ with a familiar group?

September 12th, 2017, 11:33 AM  #10 
Member Joined: May 2017 From: Russia Posts: 34 Thanks: 5 
The quotient group $\displaystyle F_{p(p1)}/F'_{p(p1)}$ consists of (p1) congruence classes $\displaystyle aH,\ a\in \mathbb{F}_p, \ H=F'_{p(p1)}$. The classes are $\displaystyle \begin{pmatrix} a&0 \\\ 0&1\end{pmatrix}H=\left\{ \begin{pmatrix} a&b \\\ 0&1\end{pmatrix} \ \bigg \ b\in\mathbb{F}_p \right\}$. The operation in the quotient group: $\displaystyle \begin{pmatrix} a_1&0 \\\ 0&1\end{pmatrix}H \cdot \begin{pmatrix} a_2&0 \\\ 0&1\end{pmatrix}H = \left(\begin{pmatrix} a_1&0 \\\ 0&1\end{pmatrix}\cdot\begin{pmatrix} a_2&0 \\\ 0&1\end{pmatrix}\right)H=\begin{pmatrix} a_1a_2&0 \\\ 0&1\end{pmatrix}H $ The quotient group is isomorphic to the multiplicative group of field $\displaystyle \mathbb{F}_p$: $\displaystyle \varphi_2: a \mapsto \begin{pmatrix} a&0 \\\ 0&1\end{pmatrix}H, \ \ a\in \mathbb{F}_p, \ a\neq 0. $ 

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