1. ## Frobeniusgroup

Let $\displaystyle F_{20}$ be the Frobeniusgroup of order $\displaystyle 20.$

1) Show that $\displaystyle F_{20}$ has the trivial Center.
2) Determine for every divisor $\displaystyle d$ of $\displaystyle 20$ the number of elements in $\displaystyle F_{20}$ of order $\displaystyle d.$
3)Determine the number of 5-Sylowgroups in $\displaystyle F_{20}$ and teh number of 2-Sylowgroups in $\displaystyle F_{20}$ .

4) Determine the normal subgroups of $\displaystyle F_{20}$ .

2. The Frobeniusgroup of order 20
is the set of maps from $\displaystyle F_5$ to $\displaystyle F_5$ of the form
$\displaystyle x \mapsto ax+b$ where $\displaystyle a,b \in F_5, b \neq 0.$
slightly different order:

2)
d=1: 1 element, the identity $\displaystyle x$
d=2: 5 elements, $\displaystyle -x+b, b \in F_5$
d=4: 10 elements, $\displaystyle ax+b, \; a\in \{2,3\}, b \in F_5$
d=5: 4 elements, $\displaystyle x+b, \; b \in F_5, b \neq 0$

1)
A presentation for $\displaystyle F_{20}$ is

$\displaystyle \alpha^5 = \beta^2 = \gamma^4 = 1, \gamma^2 = \beta, \alpha^\beta=\alpha^{-1}, \alpha^\gamma=\alpha^2$

We take
$\displaystyle \alpha = x+1, \beta = -x, \gamma = 2x$
The elements of order 5 and two together lie in the
Dihedral group of generated by $\displaystyle \alpha$ and $\displaystyle \beta$ (of order 10).
This group has trivial center. Hence $\displaystyle F_{20}$ has trivial center.

3) There is only one Sylow 5 subgroup, the group
generated by $\displaystyle \alpha = x+1$ (of order 5).
There are $\displaystyle 5$ Sylow 2 subgroups (of order 4).
They are the groups generated by $\displaystyle \gamma = 2x$ and its conjugates
$\displaystyle 2x+b \; (b \neq 0)$.

4) There are two normal subgroups, both mentioned already:
The Sylow 5 subgroup generated by $\displaystyle \alpha$,
and the Dihedral group of order 10 generated by $\displaystyle \alpha$ and $\displaystyle \beta.$

Best,

ZD