Groups acting on groups

Question

When a group acts on a set, many doors open up. The orbits partition the set, the stabilizers are subgroups, the size of an orbit is the index of the stabilizer, the # of orbits is the average # of fixed points, etc. But when a finite group acts on a finite-dimensional complex vector space, the additional structure reveals even more information about the action. For example, the representation decomposes into irreducible subrepresentations that can be described in a character table.

I would suspect that if a group $G$ acts on a group $H$ in a sensible way, then one can derive additional information beyond the possibilities of a group acting on a set, just like in representation theory. By "sensible way" I mean that the action satisfies the additional property $g * (h h') = ( g * h) (g * h')$ for all $g \in G$ and $h, h' \in H$, which gives a group homomorphism $G \to \mathrm{Aut}(H)$. One example is $G$ acting on itself by conjugation. If $H$ is an abelian group, then 2021 Wikipedia would call this sort of action (together with $H$) a $G$-module, although I think there are different uses of that term.

One consequence of the new property is that
$$g * 1_H = g * (1_H 1_H) = (g * 1_H)(g * 1_H) \in H \text, $$
so $g * 1_H = 1_H$ for all $g \in G$; $1_H$ is always a fixed point of the action.

What are some properties of groups acting on groups that are not always true of groups acting on sets?

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Imposing too many constraints can make the phenomenon bland. If $g * (h h') = (g * h)h'$ for all $g \in G$ and $h, h' \in H$, then $g * h = g * (1_H h) = (g * 1_H)h = 1_H h = h$ for all $h \in H$, so the action is trivial. This is related to Notions of groups acting on groups, although that question relinquishes the assumption that $g * (h h') = (g * h) (g * h')$.

Answer 1

If a group $G$ acts on another group $H$ (by automorphisms), then a distinguishing condition takes place, namely:

$$\operatorname{Fix}(g)\le H, \forall g\in G \tag 1$$

where $\operatorname{Fix}(g):=\{h\in H\mid \phi_g(h)=h\}$, being $\phi\colon G\longrightarrow \operatorname{Aut}(H)$ the action. In fact, for every $g\in G$, $\phi_g(1_H)=1_H$, whence $1_H\in\operatorname{Fix}(g)\ne\emptyset$; moreover, by definition, $\operatorname{Fix}(g)\subseteq H$; finally, for every $g\in G$ and $h_1,h_2\in\operatorname{Fix}(g)$, $\phi_g(h_1h_2^{-1})=$ $\phi_g(h_1)\phi_g(h_2^{-1})=$ $\phi_g(h_1)\phi_g(h_2)^{-1}=$ $h_1h_2^{-1}$, whence $h_1h_2^{-1}\in\operatorname{Fix}(g)$.

If $G$ and $H$ are both finite, say $G=\{g_1,\dots,g_{|G|}\}$ and $H=\{h_1,\dots,h_{|H|}\}$, then the usual equation:

$$\sum_{i=1}^{|H|}|\operatorname{Stab}(h_i)|=\sum_{j=1}^{|G|}|\operatorname{Fix}(g_j)| \tag 2$$

and the usual condition (by Burnside's Lemma):

$$|G|\mid \sum_{j=1}^{|G|}|\operatorname{Fix}(g_j)| \tag 3$$

give new opportunities, according to the fact that -by Lagrange's Theorem- the summands in the RHS of $(2)$ and in $(3)$ must all divide $|H|$.

As an entry test for this new setting, let's take $G=C_p$ and $H=C_q$, where $p$ and $q$ are distinct primes. If a nontrivial homomorphism exists, then $|\operatorname{Fix}(g_{\bar j})|=1$ and $|\operatorname{Stab}(h_{\bar i})|=1$, for some $\bar j\in \{1,\dots,p\}$, $\bar i\in\{1,\dots,q\}$. Then $(2)$ and $(3)$ yield:

$$[\space k+(q-k)p=l+(p-l)q\Longrightarrow k(p-1)=l(q-1)\space]\space\wedge\space [\space p\mid pq-l(q-1)\space ] \tag 4$$

for some $1\le k\le q$ and $1\le l\le p$. Now:

• if $p>q$, then from $(4)$-2nd term of the "$\wedge$": $p\mid pq-l(q-1)\Longrightarrow$ $p\mid l \Longrightarrow$ $l=p\Longrightarrow$ $k(p-1)=p(q-1)\Longrightarrow$ $k=\frac{p}{p-1}(q-1)>q-1\Longrightarrow$ $k=q$; but $(k,l)=(q,p)$ is not a solution of $(4)$-1st term of the "$\wedge$": so, for $p>q$, there are no nontrivial homomorphisms $\phi\colon C_p\longrightarrow\operatorname{Aut}(C_q)$;
• if $p<q$ and $p\nmid q-1$, then from $(4)$-2nd term of the "$\wedge$": $p\mid pq-l(q-1)\Longrightarrow$ $p\mid l$, and we fall back into the previous case.

Therefore, if $p\nmid q-1$, there are no nontrivial homomorphisms $\phi\colon C_p\longrightarrow\operatorname{Aut}(C_q)$. This hasn't used any knowledge about the isomorphism class of the group $\operatorname{Aut}(C_q)$. (Incidentally, that for $p\mid q-1$ there is actually a nontrivial homomorphism $\phi$, it is shown e.g. here.)

As one more application, let's now assume $G=C_p$ and $|H|=q^2$, still with $p,q$ distinct primes. If a nontrivial homomorphism exists, then $(2)$ and $(3)$ yield:

$$[\space k+(q^2-k)p=l_0+l_1q+(p-l_0-l_1)q^2\space]\space\wedge\space [\space p\mid k\space ] \tag 5$$

for some $1\le k\le q^2$ and $1\le l_0+l_1\le p$. Since $p\mid k\Longrightarrow p\le k\le q^2$, for $p>q^2$ there are no nontrivial homomorphisms $\phi$. This suffices to give a different proof to, e.g., this question (and the conclusion doesn't change with $\Bbb Z_2\times \Bbb Z_2$ in place of $\Bbb Z_4$). Again, this hasn't used any knowledge about the isomorphism class of the group $\operatorname{Aut}(H)$.