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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.

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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".

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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
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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.

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