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 September 11th, 2017, 06:29 PM #1 Member   Joined: Jan 2015 From: usa Posts: 92 Thanks: 0 center of a goup Let $p$ be a prime number. We define the frobenius group by $F_{p(p-1}=\left\{\begin{pmatrix}a&b\\0&1\end{pmatrix}, a\in \mathbb{F}_p^{\times}, b\in \mathbb{F}_p\right\}$ I want to identify the center $Z(F_{p(p-1)})$ and $F'_{p(p-1)}=[F_{p(p-1)},F_{p(p-1)}]$ with I will be gratefull if you could help me.
September 11th, 2017, 07:09 PM   #2
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Quote:
 Originally Posted by mona123 Let $p$ be a prime number. We define the frobenius group by $F_{p(p-1}=\left\{\begin{pmatrix}a&b\\0&1\end{pmatrix}, a\in \mathbb{F}_p^{\times}, b\in \mathbb{F}_p\right\}$ I want to identify the center $Z(F_{p(p-1)})$ and $F'_{p(p-1)}=[F_{p(p-1)},F_{p(p-1)}]$ with I will be gratefull if you could help me.
What's the square bracket notation?

 September 11th, 2017, 09:06 PM #3 Senior Member   Joined: Aug 2012 Posts: 1,777 Thanks: 482 I convinced myself that the center must be the identity matrix and nothing else. Let $u, v \in \mathbb F_p^\times$ and $n, m \in \mathbb F_p$. Let $\langle u, n \rangle$ be the matrix whose top row contains $u$ and $n$ and bottom row is $0$ and $1$. This is just a notational device to save me the trouble of writing out the matrices. Now multiplying $\langle u, n \rangle$ by $\langle v, m \rangle$ on both the right and on the left, we see that for the product matrices to be identical we must have (*) $m(u - 1) = n(v - 1)$. (I left out all the detail in this step, you need to multiply out the matrices to see why this is so). Now suppose the $\langle u, n \rangle$ matrix is in the center. That means it must commute with EVERY $\langle v, m \rangle$. Clearly if $\langle u, n \rangle= \langle 1, 0 \rangle$ then (*) is true for any $\langle v, m \rangle$ so that $\langle 1, 0 \rangle$ is in the center. But that's just the identity matrix so this should not be a surprise. Now we have to show that $\langle 1, 0 \rangle$ is the ONLY matrix that commutes with everything. We have to prove that if either $u \ne 1$ or $n \ne 0$ then SOME matrix fails to commute. Now if $u \ne 1$ then we can take $m \ne 0$ to make the left side of (*) nonzero. Then we can take $v = 1$ to make the rhs of (*) zero and commutativity is violated. On the other hand suppose $n \ne 0$. Then we can take $v = 0$ to make the rhs of (*) nonzero; and taking $m = 0$ makes the lhs of (*) zero, so it doesn't commute. Therefore the center consists only of the identity. **** Sorry wait I have an error I have to fix. If my edit window runs out, stay tuned. $v$ can't be $0$ because $v$ is a unit. **** What I need in the $n \ne 0$ case is for the rhs $n(v - 1)$ to be nonzero. So $v - 1$ has to be nonzero. If $p > 2$ we can take $v = 2$ and we're good. If $p = 2$ then this is a little fuzzy. It looks like in this case then BOTH elements are in the center. Is that right? I'll leave this to you. ps -- https://en.wikipedia.org/wiki/Frobenius_group says that $p > 2$. Is that given? They said (using a different definition) that the smallest Frobenius group has order $6$, meaning that $p = 3$. If $p = 2$ then the Frobenius group is just $\mathbb Z_2$ which is Abelian, so the whole group is indeed the center. This must be a degenerate case of the idea described in the Wiki page, which describes these groups in a completely different but presumably equivalent way. Last edited by Maschke; September 11th, 2017 at 09:55 PM.

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