Actions and Sylow theorems · Definition

Group action

modern algebragroups

Before you start you need to know

Actions let a group move the elements of a set, and they are how groups appear as symmetries.

Definition

An action of a group \(G\) on a set \(X\) assigns to each \(g,h\in G\) and \(x\in X\) an element \(g\cdot x\in X\), so that the identity \(e\) moves nothing and the product of \(G\) acts by composition.

\[e\cdot x=x,\qquad (gh)\cdot x=g\cdot(h\cdot x)\]

The prototypical example is \(S_n\) acting on \(\{1,\dots,n\}\) by \(\sigma\cdot i=\sigma(i)\); the dihedral group acts on the vertices of a polygon, and any group acts on itself by left multiplication, \(g\cdot x=gx\), or by conjugation, \(g\cdot x=gxg^{-1}\).

From the two axioms it follows that each \(g\) induces a bijection \(x\mapsto g\cdot x\) of \(X\), with inverse \(x\mapsto g^{-1}\cdot x\). So an action is the same as a homomorphism from \(G\) to the group of permutations of \(X\), which connects this concept with \(S_n\) and with Cayley's theorem.

It is important to respect the order in \((gh)\cdot x=g\cdot(h\cdot x)\): first \(h\) acts and then \(g\). Formulas such as \(g\cdot x=xg\) fail this axiom in non-abelian groups. Next, actions are used to group the elements of \(X\) into orbits and to count.

Think about it

Is \(g\cdot x=xg^{-1}\) an action of \(G\) on itself?

Yes: \(e\cdot x=x\) and \(g\cdot(h\cdot x)=xh^{-1}g^{-1}=x(gh)^{-1}=(gh)\cdot x\).

Exercises 10

References

  • David S. Dummit, Richard M. Foote, Abstract Algebra, 3rd edition, Wiley, 2004. §1.7
  • John B. Fraleigh, A First Course in Abstract Algebra, 7th edition, Pearson, 2002. §16