Group Actions
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Introduction
Historically, groups originated from the study of Galois theory as the action of symmetric groups on roots of polynomials. In a more “modern” context, groups play a significant roles in many other areas of study. As it turns out, we can understand quite a bit about groups by investigating how they act on various algebraic structures. As such, we are very interested as a whole about how groups act.
Group Actions 101
Our goal will be to eventually think of a groups’s elements as functions (specifically, permutations). We run into the immediate issue that an abstract group’s elements are not necessarily functions; however, we see will see that we can reduce to this case by imposing certain axioms.
[definition] [deftitle]Definition (Group Action).[/deftitle]
A (left) group action of a group $G$ on a set $A$ is a map
\[\begin{align*} G\times A &\to A \\ (g, a) &\mapsto g\cdot a \end{align*}\]that satisfies the properties:
- $g \cdot (h \cdot a) = (gh)\cdot a$ for all $g, h\in G$ and $a \in A$; and
- $1 \cdot a = a$ for all $a \in A$. [/definition]
It is a huge pain in the butt to keep writing $g\cdot a$, so we will abuse notation and write
\[ga \coloneqq g\cdot a\]instead when there is no risk of confusion of group multiplication and the action. For convenience, we will also say that $A$ is a $G$-set to mean that $G$ acts on $A$.
[example] [extitle]Remark.[/extitle]
Property 1, along with the associativity property of groups, imply that group actions obey the associativity property of functions (if we think of the group elements as functions). The fact that every element in a group is invertible means there is an inverse function for each group element. Furthermore, Property 2 guarantees that a functional identity function exists so it actually makes sense to talk about inverse functions. [/example]
We are now going to make this remark precise. Let $A$ be a $G$-set. Then, for each $g \in G$, we define a map
\[\begin{align*} \sigma_g : A &\to A \\ a &\mapsto ga. \end{align*}\]It is pretty immediate that the map $G \to \Sym_A$ given by $g \mapsto \sigma_g$ is a group homomorphism. And, so, we get the following property:
[theorem] [thmtitle]Theorem.[/thmtitle]
Each $\sigma_g$ is a permutation of $A$. [/theorem]
[proof] Since $g \mapsto \sigma_g$ is a homomorphism, it follows that
\[\sigma_{g^{-1}} \circ \sigma_g = \sigma_{g^{-1}g} = \sigma_1 = \sigma_{gg^{-1}} = \sigma_{g} \circ \sigma_{g^{-1}}.\]Thus, each $\sigma_g$ has a two-sided inverse $\sigma_{g^{-1}}$ and we are done. [/proof]
Notice that what this hints at is that group actions can be understand from the perspective of permutations. In fact, group actions can be fully characterized in this way by the following theorem.
[theorem] [thmtitle]Theorem (Group actions are homomorphisms into the symmetric group).[/thmtitle]
There is a 1-1 correspondence
\[\{G\text{-actions on } A\} \leftrightarrow \Hom(G, \Sym_A).\][/theorem]
[proof] The work we did above proves that there is a homomorphism $G \to \Sym_A$ for a particular group action of $G$ on $A$. We claim that there is an inverse map that takes $\phi \in \Hom(G, \Sym_A)$ to a group action of $G$ on $A$. Define the map
\[\begin{align*} G\times A &\to A \\ g\cdot a &\mapsto \phi(g)(a). \end{align*}\]It is clear that this is a $G$-action on $A$ and this map is the inverse that we are looking for. [/proof]
This theorem justifies the following terminology:
[definition] [deftitle]Definition (Permutation Representation).[/deftitle]
Given a group action $G$ on the set $A$, the associated group homomorphism $\phi: G \to \Sym_A$ is called the permutation representation of the group action. Furthermore, the kernel of the group action is $\ker\phi$. If $\ker\phi$ is trivial, then the group action is said to be faithful. [/definition]
Basic Examples of Group Actions
[example] [extitle]Trivial action.[/extitle]
Let $G$ be a group and $A$ a nonempty set. Then the group action $ga\coloneqq a$ for all $g\in G$ and $a \in A$ is known as the trivial action. The associated permutation representation sends every $g \in G$ to the identity permutation. This is as un-faithful of a group action as you can get (provided $|G| > 1$). [/example]
[example] [extitle]Natural symmetric group action.[/extitle]
Let $A$ be a nonempty set. Then $\Sym_A$ acts on $A$ in a natural way: $\sigma a \coloneqq \sigma(a)$ for all $\sigma\in \Sym_A$ and $a \in A$. The associated permutation representation sends each $\sigma$ to itself and, so, the natural action is a faithful one. [/example]
[example] [extitle]Action of $\Sym_n$ on ${[n] \choose k}$.[/extitle]
Given any subset
\[\{n_1, \ldots, n_k\} \subseteq [n],\]there is a natural action of $\Sym_n$ defined by
\[\sigma \cdot \{n_1, \ldots, n_k\} \coloneqq \{\sigma(n_1), \ldots, \sigma(n_k)\}.\]This generalizes the natural symmetric group action by restricting to the case where $k = 1$. [/example]
[example] [extitle]Action of $\Sym_n$ on $[n]^k$.[/extitle]
Given any $k$-tuple
\[(n_1, \ldots, n_k) \in [n]^k,\]there is a natural action of $\Sym_n$ defined by
\[\sigma \cdot (n_1, \ldots, n_k) \coloneqq (\sigma(n_1), \ldots, \sigma(n_k)).\]This generalizes the natural symmetric group action by restricting to the case where $k = 1$. [/example]
[example] [extitle]Dihedral groups.[/extitle]
Label the vertices of a regular $n$-gon by $[n] \coloneq {1, \cdots, n}$. Then there is a natural faithful action of $D_{2n}$ on $[n]$ by permutation given by way the symmetries of $D_{2n}$ permute the corresponding vertices. [/example]
[example] [extitle]Left regular action.[/extitle]
Every group $G$ acts naturally on itself by left-multiplication. This is sometimes called the left regular action. That is, $g\cdot a \coloneqq ga$ for all $g, a\in G$. The cancellative property of groups implies that this action is faithful (this later remark is called Cayley’s theorem). [/example]
[example] [extitle]Left action on cosets.[/extitle]
Given a group $G$ and a (not necessarily normal) subgroup $H$ of $G$, there is a natural action of $G$ on the set of left cosets $G/H$. In particular,
\[g\cdot xH \coloneqq (gx)H\]for all $g\in G$ and $xH \in G/H$. This is a generalization of the left regular action which can be seen by taking $H = \{1\}$ and noting $G/\{1\} \cong G$. [/example]
[example] [extitle]Conjugation action.[/extitle]
Every group $G$ acts naturally on itself by conjugation. That is,
\[g\cdot a \coloneqq gag^{-1}\]for all $g, a\in G$. Cancellation implies that the conjugation action is faithful. In fact, we can say something quite a bit stronger: conjugation by $g$ is an automorphism of $G$. [/example]
[example] [extitle]Conjugation action on cosets.[/extitle]
The conjugation action can be generalized to the conjugation action on the power set $2^G$ of $A$. In particular, we define
\[g\cdot A \coloneqq gAg^{-1} \coloneqq \{gag^{-1} \mid a \in A\}\]for all $g\in G$ and $A \subseteq G$. [/example]
[example] [extitle]Group Representation.[/extitle]
If we fix a $k$-vector space $V$ and a group homomorphism $\phi:G \to \GL(V)$, then the group $G$ acts linearly on $V$ by
\[g\cdot v \coloneqq \phi(g)(v)\]for all $g \in G$ and $v \in V$. This is known as a group representation of $G$ and is a construction of major interest in representation theory. [/example]
Some Worked Exercises
[example] [extitle]Dummit and Foote Problem 1.7.10a[/extitle]
The action of $\Sym_n$ on ${[n]\choose k}$ is faithful for every $k \in [n-1]$ but not $[n]$ unless $n = 1$. [/example]
[proof] The case $n = 1$ is clear so we skip over it. For $n > 1$, let $\sigma$ be in the kernel of the action. Then we know that for any $i \in [n]$
\[\sigma \cdot ([n] \setminus i) = [n] \setminus i.\]Since $\sigma$ is a bijection of $[n]$, this implies that $\sigma(i) = (i)$. This implies that $\sigma$ is the identity permutation. Thus, the kernel of the action is trivial and we are done. [/proof]
[example] [extitle]Dummit and Foote Problem 1.7.10b[/extitle]
The action of $\Sym_n$ on $[n]^k$ is faithful for every $k \in [n]$. [/example]
[proof] Since
\[\sigma\cdot (n_1, \ldots, n_k) = (n_1, \ldots, n_k),\]it follows that $\sigma(n_i) = n_i$. Since $n_i$ is arbitrary, it follows that $\sigma$ is the identity permutation on $[n]$ and the kernel of the action is trivial. [/proof]
[example] [extitle]Dummit and Foote Problem 1.7.13[/extitle]
Assume $n$ is an even positive integer. Then
\[D_{2n} = \langle r, s \mid r^n = s^2 = srsr = 1\rangle\]acts on the set consisting of pairs of opposite vertices of a regular $n$-gon and the kernel of said action is $\{1, r^{n/2}\}$ for $n > 4$. If $n = 4$, then the kernel is $\{1, r^2, sr^2\}$. [/example]
[proof] Label the vertices of the $n$-gon by $[n]$ counterclockwise and let $k = n/2$. Then the set of opposite pairs of vertices is given by
\[X = \{\{i, k + i\} \mid i \in [k]\}.\]Then the group action of $D_{2n}$ on $X$ is given by permutations of the labels corresponding to a symmetry in $D_{2n}$. That is,
\[s^a r^b \cdot \{i, k + i\} \coloneqq \{\sigma(i), \sigma(k+i)\}\]where
\[\sigma = [(2,n)(3,n-1)\cdots(k, k+2)]^a(1, 2, \ldots, n)^b \in \Sym_n.\]It is fairly immediate that this actually defines a group action so we omit the proof of that claim and proceed to investigate the kernel. Clearly, $1$ and $r^k$ are in the kernel of the action. We show that these are the only ones that are. It is immediate that $s$ and $r^j$ are not in the kernel for $j \neq k, n$ so it remains to show that $sr^j$ is not in the kernel for $j = 0, \ldots,n - 1$. Consider
\[sr^j \cdot \{1, k+1\}.\]In order for $sr^j$ to be in the kernel, it would have to send $1$ to $k+1$ and $k+1$ to $1$, or $1$ and $k+1$ are fixed points. If $sr^j$ sends $1$ to $k+1$, then $j = k$ is forced. But this would mean that $2$ is sent to $k + 2$ then sent to $k$ and, for $n > 4$, $k\neq 2$ (but it does if $n = 4$). That is,
\[sr^j \cdot \{2, k+2\} \neq \{2, k+2\}\]unless $k = 2$ as desired. [/proof]