1. ## Frobeniusgroup

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

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

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

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

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

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

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

We take
$
\alpha = x+1, \beta = -x, \gamma = 2x
$

The elements of order 5 and two together lie in the
Dihedral group of generated by $\alpha$ and $\beta$ (of order 10).
This group has trivial center. Hence $F_{20}$ has trivial center.

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

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

Best,

ZD