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Section 2.3 Group Actions and applications

A powerful tool both within algebra and other areas of mathematics is the notion of a group action. There are two equivalent ways in which the characterize a group action. We begin with the more “constructive” one.

Definition 2.3.1.

Let \(G\) be a group and \(X\) a set. We say that \(G\) acts on the set \(X\) if there is a map \(G\times X \to X\text{,}\) denoted \((g,x)\mapsto g\cdot x\text{,}\) satisfying two properties:

  • \(e\cdot x = x \text{ for all } x\in X\text{.}\) Here \(e\) is the identity of the group \(G.\)

  • \(g_1\cdot(g_2\cdot x) = (g_1g_2)\cdot x\) for all \(g_1, g_2 \in G\) and \(x\in X.\)

Example 2.3.2.

We say that the group \(G\) acts on itself by conjugation if \(X=G\) and the map \(G\times G\to G\) is given by \((g,x)\mapsto gxg^{-1}.\)

Example 2.3.3.

Let \(G\) be a group, \(H\) a subgroup, and \(X=G/H\) the set of left cosets. We say that the group \(G\) acts on \(G/H\) by left translation if the map \(G\times G/H\to G/H\) is given by \((g,xH)\mapsto gxH.\)

Example 2.3.4.

Let \(G\) be a finite group and for a prime \(p\text{,}\) suppose that \(p^m \mid |G|\text{.}\) Let

\begin{equation*} X = \{ H \le G\mid |H|=p^m\}\text{.} \end{equation*}

The Sylow theorems tell us that \(X\) is a non-empty set. We let the group \(G\) act on \(X\) by conjugation via

\begin{equation*} (g, H) \mapsto gHg^{-1}. \end{equation*}

We note that this makes sense since for a given \(g\in G\text{,}\) the map \(\varphi_g:G\to G\) given by \(\varphi_g(x) = gxg^{-1}\) is an isomorphism, and as such

\begin{equation*} gHg^{-1} = \varphi_g(H) \end{equation*}

is a subgroup of \(G\) having the same cardinality at \(H\text{,}\) so is again in \(X.\)

There is an equivalent way in which to think of a group action \(G\) acting on a set \(X\text{,}\) and that is as a group homomorphism \(\varphi: G \to Per(X)\text{,}\) where \(Per(X)\) is the set of bijections \(X \to X\) viewed as a group under function composition.

Remark 2.3.6.

This may seem a bit complicated at first glance. Simply realize that the homomorphism \(\varphi: G \to Per(X)\) has as its codomain the group of bijective functions \(X\to X\text{.}\) These are simply bijections as in general \(X\) has no algebraic structure. On the other hand, this explains our notation: Since \(\varphi(g)\) is a function, it makes intuitive sense to name it as such, so we set \(\varphi(g) = \varphi_g\) where \(\varphi_g:X\to X\) is a bijective map.

There are very good reasons for utilizing this correspondence. Group actions viewed as maps \(G\times X \to X\) are easy to describe, but have less obvious group-theoretic implications. However, an action described as a homomorphism \(\varphi:G \to Per(X)\) has obvious algebraic objects related to it, such as its kernel.

Example 2.3.7.

Let a group \(G\) act on itself by left translation. That means there is a map \(G\times G \to G\) given by \((g,x) \mapsto gx\) where the juxtaposition \(gx\) is the product in the group.

The associated permutation representation is

\begin{equation*} \varphi: G\to Per(G) \text{ given by } \varphi(g) = \varphi_g \text{ with } \varphi_g(x) = gx\text{,} \end{equation*}

and is called the left regular representation of \(G.\)

What is the kernel of \(\varphi\text{?}\) By definition,

\begin{equation*} \ker \varphi = \{g \in G\mid \varphi(g) = \varphi_g = e_{Per(G)}\}, \end{equation*}

the identity bijection. That means that \(\varphi_g(x) = gx = x\) for all \(x\in X=G.\) Simply taking \(x=e_G\) tells us that \(g=e,\) so the kernel is trivial and this permutation representation is injective.

As a corollary of this observation we obtain Cayley's theorem.

To gain further insight into group actions, we need to define two more notions, an orbit and a stabilizer.

Definition 2.3.9.

Let \(G\) act on a set \(X\) and let \(x\in X\text{.}\)

  • The stabilizer or isotropy subgroup of \(x\) is

    \begin{equation*} G_x := \{ g\in G \mid g\cdot x = x\}\text{.} \end{equation*}

  • The orbit of \(x\) is

    \begin{equation*} Gx=G\cdot x := \{ g\cdot x\mid g\in G\} =\{y\in X\mid y = g\cdot x \text{ for some } g\in G\}\text{.} \end{equation*}

Note that \(G_x\) is a subgroup of \(G\text{,}\) while \(Gx\) is a subset of \(X\text{.}\)

Consider a couple of examples to anchor these definitions.

Example 2.3.10.

Let \(G\) act on itself by conjugation, and let \(x\in X.\) Then

\begin{equation*} Gx = \{gxg^{-1}\mid g\in G\} \end{equation*}

is the conjugacy class of \(x\text{,}\) while

\begin{equation*} G_x = \{g\in G\mid gxg^{-1} = x\} \end{equation*}

is the centralizer of \(x.\)

Example 2.3.11.

Let \(G\) be a group, and \(X =\{H \le G\}\) be the set of subgroups of \(G\text{.}\) \(G\) acts on \(X\) by conjugation. Let \(x=H \in X\text{.}\) Then

\begin{equation*} Gx = \{gHg^{-1}\mid g\in G\} \end{equation*}

is the conjugacy class of \(H\text{,}\) while

\begin{equation*} G_x = \{g\in G\mid gHg^{-1} = H\} \end{equation*}

is the normalizer of \(H.\)

Exercise 2.3.1.

Let \(G\) act by conjugation on the set \(X\) of all subgroups of \(G\) as in the second example above.

(a)

What does it mean when the size of the orbit is one, meaning the orbit consists only of \(H\text{?}\) How is an orbit of size one related to the stabilizer?

Answer.

An orbit of size one says \(H = gHg^{-1}\) for all \(g\in G,\) that is \(H\) is a normal subgroup. It follows that in this case, the isotropy subgroup (normalizer of \(H\)) is all of \(G\text{,}\) so

\begin{equation*} |Gx|=1\iff G_x = G\text{.} \end{equation*}

We shall generalize this below.

(b)

Let \(G=S_3\) and \(X\) all the subgroups of \(S_3.\) For each \(x\in X\text{,}\) compute the orbit and stabilizer.

Answer.

All proper subgroups of \(S_3\) are cyclic which makes our notation easier. Let \(e\) be the identity, \(t\) be any transposition: \((1\ 2),\) \((2\ 3)\text{,}\) or \((1\ 3)\text{,}\) and \(T\) be either 3-cycle: \((1\ 2\ 3)\) or \((1\ 3\ 2).\)

\begin{align*} x\amp =\{e\}\quad Gx=\{e\}\quad G_x = G.\\ x\amp= \la t \ra\quad Gx = \{\la(1\ 2) \ra, \la (2\ 3)\ra, \la (1\ 3)\ra\}\quad G_x = \la t\ra.\\ x\amp= \la T\ra\quad Gx = \{x\}\quad G_x=G.\\ x\amp = S_3\quad Gx=\{x\}\quad G_x=G. \end{align*}

Let's show that the map \(gG_x \mapsto gx\) is well-defined. Suppose that \(gG_x = hG_x\text{.}\) By Proposition 1.3.4, the two cosets are equal iff \(h^{-1}g \in G_x\text{,}\) the stabilizer. So \(gG_x = hG_x\) iff \(h^{-1}g x = x\) iff \(gx = hx\) (using the basic properties of a group action). Thus the map is not only well-defined but also one-to-one. It is also clear that it is surjective since given \(gx \in Gx\text{,}\) the coset \(gG_x\) maps onto it.

The final statement is immediate since

\begin{equation*} |Gx| = |G/G_x| = [G:G_x]\text{,} \end{equation*}

the last equality by definition.

Since we know that that cosets of \(G_x\) in \(G\) partition \(G,\) it should come as no surprise that the orbits partition \(X.\)

By Definition 1.2.1, we need only check that every element of \(X\) is in some orbit, and that two orbits are either disjoint or equal.

By the property of a group action that requires \(e\cdot x = x\) for all \(x\in X\text{,}\) we see that \(x \in Gx\) (no matter what \(G\) is).

Suppose that \(z \in Gx \cap Gy\text{.}\) We must show that \(Gx = Gy\text{.}\) By symmetry, it is enough to show \(Gx\subseteq Gy\text{.}\) Since \(z \in Gx \cap Gy\text{,}\) we may write

\begin{equation*} z = g_1x = g_2y\text{.} \end{equation*}

But that means that \(x = (g_1^{-1}g_2)y\text{,}\) so for any \(g\in G\text{,}\)

\begin{equation*} gx = (gg_1^{-1}g_2)y \in Gy \end{equation*}

which completes the proof.

Finally, let's prove a couple of nontrivial results with group actions. For the first, we learned early on that a subgroup \(H\) for which the index \([G:H]=2\) is normal (since there are only two (left or right) cosets one of which is \(H)\text{.}\) The following is a good deal stronger.

We use a lemma to simply the proof of the proposition, but which is useful in its own right.

The permutation representation \(\varphi\) acts by

\begin{equation*} \varphi(g) = \varphi_g\text{ where } \varphi_g(xH) = gxH. \end{equation*}

We note that \(k\in \ker \varphi\) if and only if \(\varphi(k) = \varphi_k\) is the identity map. So if \(k \in \ker \varphi\text{,}\) then

\begin{equation*} \varphi_k(xH)= kxH = xH \end{equation*}

for all \(x\in G.\) But simply choosing \(x=e\) gives us that \(kH = H\) which means \(k \in H.\)

Now we return to our proposition.

Let \(G\) be a finite group, and \(H\) a subgroup whose index, \([G:H]=p\text{,}\) is the smallest prime which divides the order of \(G.\) Let \(G\) act on \(G/H\) by left translation. By the lemma, the kernel \(K = \ker \varphi \subseteq H.\) To show that \(H\) is normal, we do it indirectly, by showing that \(H = K\) which is known to be a normal subgroup!

We know that \(K \subseteq H \subseteq G\) and that \(G/K \cong \Im \varphi \le Per(G/H)\) by the first isomorphism theorem. By Lagrange's theorem

\begin{equation*} |G/K| = [G:K] \mid [G:H]! = |Per(G/H)| = p!. \end{equation*}

By a simple homework type exercise we know that

\begin{equation*} [G:K] = [G:H][H:K] = p [H:K]. \end{equation*}

Putting things together we infer that

\begin{equation*} [H:K] \mid (p-1)!. \end{equation*}

But again by Lagrange,

\begin{equation*} [H:K] \mid |H| \mid |G|, \end{equation*}

so

\begin{equation*} [H:K]\mid \gcd(|G|, (p-1)!) =1 \end{equation*}

since \(p\) was the smallest prime dividing \(|G|.\) Since \([H:K]=1,\) \(H=K\) is a normal subgroup of \(G.\)

Another standard application of group actions provides the class equation which we use to show that every group of order \(p^2\) (\(p\) prime) is abelian. We of course know that every group of order \(p\) is cyclic, hence abelian.

As preface, given a finite group \(G\) we let \(G\) act on itself by conjugation. Thus the orbits are conjugacy classes of elements, and stabilizers are centralizers. An element \(z\in G\) lies in the center, \(Z(G)\text{,}\) of \(G\) if and only if its conjugacy class consists solely of the element \(z\text{,}\) and hence its centralizer, \(C_G(z)\text{,}\) is the entire group \(G.\) Because we know that that orbits partition the set (in this case the group \(G\)), we can write \(G\) as the disjoint union of orbits (conjugacy classes)

\begin{equation*} G = Gg_1 \sqcup Gg_2 \sqcup \cdots \sqcup Gg_t\text{,} \end{equation*}

recalling that the classes of size one correspond to the elements in the center of \(G.\)

As indicated above, we can write \(G\) as the disjoint union of orbits:

\begin{equation*} G = Gh_1 \sqcup \cdots \sqcup Gh_t \sqcup Gg_1 \sqcup Gg_2 \sqcup \cdots \sqcup Gg_r \end{equation*}

where the \(h_i\) are representatives of the conjugacy classes of size one, and the \(g_i\) represent those classes of size greater than one. As this is a disjoint union, it follows that

\begin{equation*} |G| = \sum_{i=1}^t |Gh_i| + \sum_{i=1}^r |Gg_i|. \end{equation*}

The first sum is the order of the center, while each summand in the second

\begin{equation*} |Gg_i| = [G:G_{g_i}] = [G:C_G(g_i)] \end{equation*}

by the orbit-stabilizer theorem, hence the result.

Exercise 2.3.2.

For a prime \(p\text{,}\) a \(p\)-group is a finite group whose order is a power of \(p.\)

(a)

Let \(G\) be any \(p\)-group. Show that \(G\) has a non-trivial center. Specifically show that \(p \mid |Z(G)|.\)

Answer.

This follows from the class equation:

\begin{equation*} |G| = |Z(G)| + \sum_{i=1}^r [G: C_G(g_i)], \end{equation*}

where \(g_1, \dots, g_r\) are representatives of the conjugacy classes of size greater than one. But that means that each \([G: C_G(g_i)] >1\) and since \([G: C_G(g_i)] \mid |G|\) by Lagrange, each \([G: C_G(g_i)]\) is a power of \(p\text{,}\) in particular divisible by \(p\text{.}\) Thus

\begin{equation*} |Z(G)| \equiv |G| - \sum_{i=1}^r[G:C_G(g_i)] \equiv 0\pmod p. \end{equation*}
(b)

Let \(G\) be a group, and suppose that \(G/Z(G)\) is cyclic. Show that \(G\) is abelian.

Answer.

Let’s write \(Z\) for \(Z(G).\) \(G/Z\) cyclic means that \(G/Z = \la xZ \ra\) for some \(x\in G.\) To show that \(G\) is abelian, choose \(g,h \in G\) and we shall show they commute.

By our assumption, \(gZ = x^mZ\) and \(hZ = x^nZ\) for some integers \(m,n.\) This means that \(g= x^m z_1\) and \(h = x^n z_2\) for some elements \(z_i \in Z.\) It follows that

\begin{equation*} gh = x^mz_1 x^n z_2 = z_1 x^{m+n}z_2 = z_2 x^{n+m} z_1 = z_2 x^n x^m z_1= x^nz_2 x^m z_1 = hg \end{equation*}

using that the \(z_i\) commute with all elements in the group and \(x^mx^n = x^n x^m.\)

(c)

Let \(G\) be a group of order \(p^2\text{.}\) Show that \(G\) is abelian.

Answer.

From the first exercise, we know that \(|Z(G)| = p\) or \(p^2\text{.}\) If \(|Z(G)|=p^2\text{,}\) then \(G=Z(G)\text{,}\) so is abelian. Assume to the contrary, \(|Z(G)|=p\) (so \(G\) is not abelian). But \(Z(G) \normal G\text{,}\) and since \(G/Z(G)\) has order \(p\text{,}\) it is cyclic. But the previous exercise shows that \(G\) is abelian, a contradiction.