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The question is related to Taft's Theorem about G-invariant radical complements. Let $A$ be an associative unitary finite-dimensional $K$-Algebra posessing a separable factor Algebra by ist nilradical. Suppose $G$ is a finite group such that the order of $G$ is not divisible by the characteristic of $K$ and $G$ is acting on $A$ as auto- or anti-automorphism. Taft has proven that a $G$-invariant radical complement exists. A $G$-invariant radical complement is a fix point of the set of all radical complements under the Action of $G$.

If we assume a finite field, then one can prove that the number of radical complements is a power of the characteristic of $K$. A natural question is:

Let $G$ be a finite $p'$-Group acting on a set $M$ of $p$-power order. Is there a fix Point in $M$ under $G$?

Within $S_3$ one can construct examples of a subgroup of order 2 acting on the cosets of $S_3$ Modulo another subgroup of order 2 -- which is a set containg 3 elements. Within this example no fix Point exist.

My questions are:

1.) What is so Special within the mentioned situation that a fix point exists.

2.) Is there a General result on fix Points related to the described context?

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    $\begingroup$ In your $S_3$-example, there is a fixed point? In fact, when $G$ is a $q$-group ($q$ prime) acting on a set $M$, then the number of fixed points in $M$ under $G$ is congruent to the size of $M$ modulo $q$. So when $\lvert M \rvert$ is coprime to $q$, then fixed points exists. The situation is different when $\lvert G \rvert $ is divisible by at least two different primes. An example would be a group $G$ of order $6$ acting on a set with $5$ elements, with one orbit of length $2$ and one of length $3$. $\endgroup$
    – Frieder Ladisch
    Commented Sep 3, 2018 at 13:00
  • $\begingroup$ Thanks for both comments. I thought maybe ist related to the fact that the radical complements are conjugated under the p-group $1+rad(A)$ and that $G$ is also acting on these "conjugation" elements. But I can't see/find a general argumentation for this fact. $\endgroup$
    – Sven Wirsing
    Commented Sep 3, 2018 at 13:28
  • $\begingroup$ Malcev had results of this nature too. $\endgroup$
    – Geoff Robinson
    Commented Sep 3, 2018 at 14:56

1 Answer 1

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I think the guess in your comment is correct: There is the following result of Glauberman:

Theorem (Glauberman). Let the finite group $G$ act on the finite group $N$ by automorphisms, where $(\lvert G \rvert, \lvert N \rvert ) = 1$. Let both $G$ and $N$ on a set $\Omega$, where the action of $N$ is transitive, and the compability condition $$ (\omega n) g = (\omega g) n^g \quad \text{for all } \omega\in \Omega,\, n\in N,\, g\in G $$ holds. Then $\Omega$ has a fixed point under $G$. Moreover, $C_N(G)$ acts transitively on the $G$-fixed points.

In your situation, this lemma applies with $N = 1 + \operatorname{rad}(A)$, which acts transitively on the set of complements of the radical, if I understand correctly.

Glauberman's result appears as Lemma 3.24 in Isaacs's text on finite groups or in Glauberman's original paper (Theorem 4).

Isaacs, I. Martin, Finite group theory. Graduate Studies in Mathematics 92. AMS 2008. ZBL1169.20001.
Glauberman, George, Fixed points in groups with operator groups, Math. Z. 84, 120-125 (1964). ZBL0123.02601.

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    $\begingroup$ Interestingly, the proof in these sources depends on knowing that at least one of $G$ and $N$ is solvable. This in turn is justified by quoting the Odd Order Theorem of Feit and Thompson. $\endgroup$
    – Richard Lyons
    Commented Sep 5, 2018 at 20:56
  • $\begingroup$ Yes, indeed. But in the particular situation of the question, we know that $N$ is nilpotent (in fact a $p$-group). $\endgroup$
    – Frieder Ladisch
    Commented Sep 6, 2018 at 15:42
  • $\begingroup$ Yes, fair enough. $\endgroup$
    – Richard Lyons
    Commented Sep 7, 2018 at 23:26
  • $\begingroup$ Taft has enhanced this theorem within his article Earl J. Taft, Orthogonai Conjugacies in Finite Groups, Math. Annalen 170, 37-40, 1967. $\endgroup$
    – Sven Wirsing
    Commented Sep 13, 2018 at 16:12

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