Abstract Nonsense

Crushing one theorem at a time

Review of Group Theorem: Group Actions (Pt. I Definitions and a Sharpening of Cayley’s Theorem)

Point of post: In this post we discuss the ideas of group actions and the counting arguments one can use them for.


We now move into one of the most beautiful subjects of finite group theory, the theory of group actions. Group actions occur everywhere in mathematics, most of the time without us even taking notice. Intuitively a group acts on a set by moving it’s members around in a specific way. For example, the permutation group S_3 acts on the set of labeled vertices of a triangle by moving them into a different configuration.

We shall see that group actions enable us to tell a lot about the structure of the underlying groups. In particular, it will give us the class equation.

Group Actions: The Definitions

Let G be a group and S a set. A mapping \cdot:G\times S\to S is called a G-action on S if

\text{ }

\displaystyle \begin{aligned}&\textbf{GA 1}\quad \cdot\left(g,\cdot\left(h,s\right)\right)=\ast\left(gh,s\right),\;\;\text{for all }g,h\in G\;\text{ and }s\in S\\ &\textbf{GA 2}\quad \cdot(e,s)=s\;\;\text{for all }s\in S\end{aligned}

 \text{ }

Remark: It is common, and we will follows this practice, to forgo the formality and denote \cdot(g,s) by g\cdot s.

If there is a specified G-action on S we are apt to call S a G-space. The first thing that should be clear is:

Theorem: Let S be a G-space with G-action \cdot. Denote then \theta_g the function S\to S given by s\mapsto g\cdot s. Then, the mapping \Phi:G\to \text{Perm}\left(S\right):g\mapsto \theta_g is a homomorphism. Moreover, any homomorphism \Psi:G\to\text{Perm}\left(S\right) induces a G-action on S given by g\cdot s=\left(\Psi(g)\right)(s).

Proof: Although obvious it is a good idea to indeed verify that \theta_g is indeed a permutation on S. To see that \theta_g is injective we merely note that if g\cdot s=g\cdot s' then g^{-1}\cdot\left(g\cdot s\right)=g^{-1}\left(g\cdot s'\right) and by \textbf{GA 1} this implies that \left(g^{-1}g\right)\cdot s=\left(g^{-1}g\right)\cdot s' or e\cdot s=e\cdots s' and by \textbf{GA 2} we are now able to conclude that s=s'. To see that \theta_g is surjective we merely note that for any s\in S we have that g^{-1}\cdot s\in S and

\text{ }

g\cdot \left(g^{-1}\cdot s\right)=\left(gg^{-1}\right)\cdot s=e\cdot s=s

 \text{ }

where we’ve used the same logic as in proving injectivity. Thus, \theta_g\in\text{Perm}\left(S\right). But, then the fact that \Phi is a homomorphism becomes trivial since for any g,h\in G and s\in S we have that

\text{ }

\left(\Phi(g)\circ\Phi(h)\right)(s)=\Phi_g\left(h\cdot s\right)=g\cdot\left(h\cdot s\right)=(gh)\cdot S=\left(\Phi(gh)\right)(s)

\text{ }

and thus \Phi(g)\circ\Phi(h)=\Phi(gh) as required.

\text{ }

Conversely, let \cdot:G\times S\to S be given by g\cdot s=\left(\Psi(g)\right)(s). Then, since \Psi is a homomorphism we have that

\text{ }

g\cdot(h\cdot s)=\left(\Psi(g)\right)\left(\left(\Psi(h)\right)(s)\right)=\left(\Psi(g)\circ \Psi(h)\right)(s)=\left(\Psi(gh)\right)(s)=(gh)\cdot s

 \text{ }

so that \cdot obeys \textbf{GA 1}. Also, since \Psi is a homomorphism we must have that \Psi(e)=\text{id}_S so that e\cdot s=\left(\Psi(e)\right)(s)=\text{id}_S(s)=s for every s\in S, and so \textbf{GA 2} is also satisfied. The conclusion follows. \blacksquare

Remark: It’s clear from the fact that for g\in G we have that \theta^{-1}_g=\theta_{g^{-1}} and so in particular if g\cdot s=s' then g^{-1}\cdot s'=s.

If S is a G-space with G-action \cdot and the associated homomorphism (as described in the last theorem) is injective we call the action \cdot faithful. We define the kernel of the G-action \cdot to be the kernel of the associated homomorphism. More explicitly if the kernel of \cdot is defined to be the set

\text{ }

\left\{g\in G:g\cdot s=s\;\;\text{for all }s\in S\right\}

 \text{ }

It’s clear from our alternate formulation of the kernel of the action that it is a normal subgroup of G.

For s\in S we define the isotropy subgroup of s, denoted G_s, to be the set

\text{ }

G_s=\left\{g\in G:g\cdot s=s\right\}

\text{ }

In other words G_s is the set of all elements of G which fix s. We now prove that the isotropy subgroup is worthy of its name:

Theorem: Let S be a G-space with G-action \cdot. Then, for any s\in S we have that G_s\leqslant G.

Proof: Let g,h\in G_s we note then that

\text{ }

(gh^{-1})\cdot s=g\cdot(h^{-1}\cdot s)=g\cdot s=s

\text{ }

and thus gh^{-1}\in G_s. The conclusion follows. \blacksquare

Remark: Notice that we used the fact that \theta_{h^{-1}}=\theta^{-1}_h.

Lastly, if S is a G-space with G-action \cdot then for s\in S we define the orbit of s, denoted \mathcal{O}_s, to be the set

\text{ }

\mathcal{O}_s=\left\{g\cdot s:g\in G\right\}

\text{ }


1.  Lang, Serge. Undergraduate Algebra. 3rd. ed. Springer, 2010. Print.

2. Dummit, David Steven., and Richard M. Foote. Abstract Algebra. Hoboken, NJ: Wiley, 200


January 4, 2011 - Posted by | Algebra, Group Theory | , , , ,


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  11. Is there a standard technical name for the following concept of defining subsets of a group by “partial specification” of their actions?

    For simplicity, consider the action of a group on itself. A group element g is thus identified as a function consisting of ordered pairs of elements of the group. Let S be a set of ordered pairs of group elements that don’t specify a many-to-one relationship. S may not contain enough ordered pairs to completely specify a function from the group onto itself. For example if the elements of the group are {a,b,c,d} the set S could be the ordered pairs { (a,b),(c,a) } which would not specify what a particular function had for (b,?) and (d,?).

    It is possible to define a product operation on such specifications by saying that the product specification R = (S)(T) is set of ordered pairs that we can form by composing the maps defined by S and T insofar as that is possible with the ordered pairs they contain. Since it’s possible to do some algebra with partial specifications, I suspect they have already been studied, but I don’t recognize a standard topic in elementary group theory that fits.

    Comment by Stephen Tashiro | October 2, 2012 | Reply

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