A question about Frobenius action

Question

On page 177 of my textbook Finite Group Theory by I. Martin Isaacs, it says:

Let $A$ and $N$ be finite groups, and suppose that $A$ acts on $N$ via automorphisms. The action of $A$ on $N$ is said to be Frobenius if $n^a\not=n$ whenever $n\in N$ and $a\in A$ are nonidentity elements. Equivalently, the action of $A$ on $N$ is Frobenius if and only if $C_N(a)=1$ for all nonidentity elements $a\in A$ and also if and only if $C_A(n)=1$ for all nonidentity elements $n\in N$.

I think I understand the “$C_A(n)$” part: $\forall n\in N$, $C_A(n)=1$ just means $\forall n\in N$, $\{a\in A\vert n^a=n\}=\{1\}$.

But I got confused when it comes to the “$C_N(a)$” part. It only assumed that $A$ acts on $N$, but not $N$ acts on $A$. If $N$ doesn’t act on $A$, how can we use the terms like “$C_N(a)$”?

If it were the action by conjugation, it would make sense. It’s easy to find that “$\forall n\in N$, $C_A(n)=\{a\in A\vert a^{-1}na=n\}=1\Leftrightarrow \forall a\in A$, $C_N(a)=\{n\in N\vert n^{-1}an=a\}=1$”. But here the “$a$” in “$n^a$” just represents an arbitrary abstract automorphism induced by $a$, not a specific one.

What have I missed? Thanks!

Answer 1

So we are given the following:

• $A$ and $N$ finite groups;
• a homomorphism $\varphi\colon A \to \operatorname{Aut}(N)$;

• an action $A \times N \to N$, defined by $(a,n) \mapsto \varphi_a(n)$

  In fact:

  a) $(e_A,n)=\varphi_{e_A}(n)=\iota_N(n)=n, \forall n \in N$;

  b) $(b,(a,n))=(b,\varphi_a(n))=\varphi_b(\varphi_a(n))=(\varphi_b\varphi_a)(n)=\varphi_{ba}(n)=(ba,n), \forall a,b \in A, n \in N$.

The stabilizer of $n \in N$ is: $\operatorname{Stab}(n)=\{a \in A \mid \varphi_a(n)=n\}$, so if the action is Frobenius, then $\forall n\in N, \operatorname{Stab}(n)=\{e_A\}$, and vice versa.

The set of elements of $N$ fixed by $a \in A$ is: $\mathcal{Fix}(a)=\{n \in N \mid \varphi_a(n)=n\}$, so if the action is Frobenius, then $\forall a\in A, \mathcal{Fix}(a)=\{e_N\}$, and vice versa.

Now, $\operatorname{Stab}(n)$ is your "$C_A(n)$" and $\mathcal{Fix}(a)$ your "$C_N(a)$".

Note that, since $|\operatorname{Stab}(n)|=1, \forall n\in N$, by the Orbit-Stabilizer Theorem we get: $|O(n)|=|A|, \forall n \in N$, and then $|\mathcal{O}|=|N|/|A|$, where $\mathcal{O}$ is the set of action's orbits. Therefore, seemingly $A$ can't act "Frobenius-like" by automorphisms on a proper subgroup of its.

Answer 2

Let $g$ be an element of a group $G$ and $X$ a subset of $G$. Define the centralizer of $g$ in $X$ by $$C_X(g) = \{\, x \in X \mid gx = xg \,\}.$$ (This definition is slightly general than the definition given by Isaacs in p. 339. Be careful; $C_X(g)$ depends on the ambient group $G$ as well which is absent.) Then the (relative) holomorph provides a way to deal with your problem uniformly. Let $H = \operatorname{Hol}(N, A) = A \ltimes N$. Consider $A$ and $N$ as subgroups of $H$. Then $$ C_A(n) = \{\, a \in A \mid na = an \,\}, \quad C_N(a) = \{\, n \in N \mid an = na \,\} $$ in $H$. Precisely, what we meant in the above is $$ an \leftrightarrow (a, 1)(1, n) = (a, n), \quad na \leftrightarrow (1, n)(a, 1) = (a, n^a). $$ So one may just write $$ C_A(n) = \{\, a \in A \mid n^a = n \,\}, \quad C_N(a) = \{\, n \in N \mid n^a = n \,\}, $$ which is apparently inconsistent notation.