frobenius group

Question

hi everyone. I want to know some example of frobenius groups. I know S_3 is a frobenius group, and if we have a semidirect produt of 2 groups then we have frobenius groups. but I want more examples of frobenius groups. Are dihedral groups , frobenius? thanks

Answer 1

Just to reiterate standard facts about Frobenius groups: a finite group $G$ is said to be a Frobenius group if $G$ contains a proper non-identity subgroup $H$ such that $H \cap gHg^{-1} = 1$ for all $g \in G \backslash H.$ This hypothesis is exactly equivalent to the permutation-group-theoretic definition of Frobenius groups: these are finite permutation groups $G$ in which no non-identity element fixes more than one point of a given set $\Omega$ on which $G$ acts transitively. Given such an action on a set $\Omega,$ just take $H$ to be a point stabilizer. Conversely, given a subgroup $H$ with $H \cap gHg^{-1} = 1$ for all $g \in G \backslash H,$ then the usual action of $G$ by right translation on the set $\Omega$ of right cosets of $H$ in $G$ gives a permutation action in which no non-identity element fixes more that one point. The fundamental theorem of Frobenius is that given a Frobenius group $G$ with such a subgroup $H,$ there is a normal subgroup $K$ with $G = KH$ and $K \cap H = \{1 \}.$ This translates the problem of finding Frobenius groups to one of finding finite groups $K$ which admit a group of automorphisms $H$ such that no non-identity element of $H$ fixes any non-identity element of $K.$ The subgroup $H$ is often known as a Frobenius complement, and the subgroup $K$ is known as a Frobenius kernel. A deep theorem of J.G. Thompson states that Frobenius kernels are nilpotent. It was known to W. Burnside that a Sylow $p$-subgroup of a Frobenius complement contains no elementary Abelian subgroup of order $p^{2}.$ Perhaps the canonical example of a Frobenius group is the case that $K$ is the additive group of a finite field with $q$ elements, and $H$ is the multiplicative group of that field, which acts by multiplication (within the field) on $K$ (when $K$ is identified with the field). Hence for any prime power $q,$ there is a Frobenius group $G$ of order $q(q-1),$ your example $S_{3}$ being one manifestation of this. An example of a Frobenius group which is not solvable occurs with $H \cong {\rm SL}(2,5)$ and $K$ Elementary Abelian of order $121$, using the action of $H$ on $K$ given by the fact that $H$ is isomorphic to a subgroup of ${\rm SL}(2,11).$

Answer 2

You can find examples and a lot of references in the survey:

A.I.Starostin, On Frobenius groups. Ukrainian Mathematical Journal, 1971, Vol. 23, No. 5, pp. 518-526.

Answer 3

the Dihedral group $D_{2n}$, when $n$ is odd, is a Frobenius group. Also if $F$ be a finite field, $H=(F\setminus\{0\},.)$ and $K=(F,+)$, then $K\rtimes H$ is a Frobenius group.

Answer 4

Frobenius groups have trivial center. This is helpful for non-examples, e.g., to see that $p$-groups, or the quaternion group is not Frobenius (and all nilpotent groups). Further examples are: non-abelian groups of order $pq$, the group $A_4$. Further non-examples: the groups $S_n$ for all $n\ge 4$.

In addition to the other answers, here is a nice discussion on Frobenius groups by Terence Tao in his tag archive: http://terrytao.wordpress.com/tag/frobenius-groups/.