Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $G$ be a finite group. We say a non-trivial group of automorphism $A$ on $G$ is regular, if each non-trivial automorphism of $A$ is regular, i.e. fixes only the identity. It is remarked in Gorenstein's Finite Groups, that the semidirect product $GA=G\rtimes A$ is a Frobenius group, and I'm having a bit of trouble showing this.

I believe that you can let $GA$ act on the set of cosets of $A$, and show that is satisfies the conditions of being a Frobenius group (transitive action, some non-trivial elements fix a letter, some non-trivial elements fix no letter, only the identity fixes more than one letter).

I don't think this should be very complicated, but the solution eludes me.

share|improve this question

2 Answers 2

Here's a hint: let $GA$ act on the elements of $G$ by the formula $$ (g,\alpha)(h) \;=\; g\,\alpha(h). $$ It shouldn't be too hard to show the required properties.

By the way, there's a certain way that you ought to be thinking about this example. If $G$ is the real numbers under addition, then the function $\alpha(x) = mx$ is an automorphism of $G$ for each nonzero $m\in\mathbb{R}$, so the group $A$ of nonzero real numbers under multiplication is a regular group of automorphisms of $G$. In this case, $GA$ is isomorphic to the group of all affine-linear functions $f(x) = b+mx$ with $m\ne 0$. This is an (infinite) Frobenius group acting on $\mathbb{R}$, with action defined by $$ (b,m)(x) \;=\; b+mx. $$ In general, elements of $GA$ can be thought of as "affine functions" on $G$, with the elements of $A$ being the possible "slopes".

share|improve this answer
I actually figured it out myself before I saw your answer. I'll add my solution as an answer. Thinking of Frobenius groups geometrically doesn't help me; I'm much more at home in finite group theory than other areas. I'll look into your example though. Thank you. :) –  Ske May 24 '12 at 16:31
Apparently, as I'm a new user, my answer will have to wait five hours. –  Ske May 24 '12 at 16:42
up vote 0 down vote accepted

Let $K$ be the finite group and $H$ the regular group of automorphisms. Then $K$ will be the Frobenius kernel and $H$ the complement in the resulting Frobenius group.

Let $G=K\rtimes H$. If we identify $K$ and $H$ with their images in $G$, we can write $G=KH$, such that the assumption of regularity becomes $k^h\neq k$ for $k\in K^*$ and $h\in H^*$. Let $G$ act by left multiplication on the set of left cosets of $H$. This is clearly a transitive action, as $g_1H$ is sent to $g_2H$ by $g_2g_1^{-1}$.

As $G=KH$, each coset can be represented by an element of $K$. If $k_1H=k_2H$, then $k_2^{-1} k_1 \in H$, but $H\cap K=1$ so $k_1=k_2$. Thus each coset of $H$ can be represented by a unique element of $K$.

If $k'\in K$ fixes a coset $kH$, then $k'kH=kH$, but then $k'k=k$ is the unique representative in $K$, so $k'=1$. Each element of $K^*$ thus fix none of the cosets of $H$. The elements of $H$ of course fix the coset $H$ itself. To see that $G$ is a Frobenius group, it therefore suffices to show that no non-trivial element of $G$ fixes more than coset.

Assume that $g\neq 1$ fixes both $kH$ and $k'H$. Then $kH=gkH$, so $k^{-1} gk$ is an element $h$ of $H$. As $g\neq 1$, we also get that $h\neq 1$. Let $k''=k^{-1} k'$. We see that $$hk''h^{-1} H=hk''H=k^{-1} gk k^{-1} k'H=k^{-1} g k' H=k^{-1} k' H=k'' H.$$

As $K$ is normal in $G$ by construction, so $hk''h^{-1}\in K$, but we saw that each coset of $H$ has a unique representative in $K$, so $hk''h^{-1}=k''$. By assumption, this implies that either $h$ or $k$ is 1, but we know that $h\neq 1$, so $1=k''=k^{-1} k'$, or rather, $k=k'$. Thus, if a $g$ fixes two coset $kH$ and $k'H$, thus coset must coincide. This proves that $G$ is a Frobenius group.

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.