maths.free › Abstract Algebra › Group Actions › Groups Acting on Sets

Groups Acting on Sets

Let X be a set and G be a group. A (left) action of G on X is a map G \times X \rightarrow X given by (g,x) \mapsto gx, where ex = x for all x \in X; (g_1 g_2)x = g_1(g_2 x) for all x \in X and all g_1, g_2 \in G

Groups Acting on Sets

Let \(X\) be a set and \(G\) be a group. A (left) action of \(G\) on \(X\) is a map \(G \times X \rightarrow X\) given by \((g,x) \mapsto gx\), where

\(ex = x\) for all \(x \in X\);
\((g_1 g_2)x = g_1(g_2 x)\) for all \(x \in X\) and all \(g_1, g_2 \in G\).

Under these considerations \(X\) is called a \(G\)-set. Notice that we are not requiring \(X\) to be related to \(G\) in any way. It is true that every group \(G\) acts on every set \(X\) by the trivial action \((g,x) \mapsto x\); however, group actions are more interesting if the set \(X\) is somehow related to the group \(G\).

Example

Let \(G = GL_2( {\mathbb R} )\) and \(X = {\mathbb R}^2\). Then \(G\) acts on \(X\) by left multiplication. If \(v \in {\mathbb R}^2\) and \(I\) is the identity matrix, then \(Iv = v\). If \(A\) and \(B\) are \(2 \times 2\) invertible matrices, then \((AB)v = A(Bv)\) since matrix multiplication is associative.

Example

Let \(G = D_4\) be the symmetry group of a square. If \(X = \{ 1, 2, 3, 4 \}\) is the set of vertices of the square, then we can consider \(D_4\) to consist of the following permutations: \[\begin{aligned}\end{aligned}\]. The elements of \(D_4\) act on \(X\) as functions. The permutation \((1 \, 3)(2 \, 4)\) acts on vertex \(1\) by sending it to vertex \(3\), on vertex \(2\) by sending it to vertex \(4\), and so on. It is easy to see that the axioms of a group action are satisfied.

In general, if \(X\) is any set and \(G\) is a subgroup of \(S_X\), the group of all permutations acting on \(X\), then \(X\) is a \(G\)-set under the group action \[\begin{aligned}\end{aligned}\] for \(\sigma \in G\) and \(x \in X\).

Example

If we let \(X = G\), then every group \(G\) acts on itself by the left regular representation; that is, \((g,x) \mapsto \lambda_g(x) = gx\), where \(\lambda_g\) is left multiplication: \[\begin{aligned}e \cdot x = \lambda_e x = ex = x \\ (gh) \cdot x = \lambda_{gh}x = \lambda_g \lambda_h x = \lambda_g(hx) = g \cdot ( h \cdot x)\end{aligned}\]. If \(H\) is a subgroup of \(G\), then \(G\) is an \(H\)-set under left multiplication by elements of \(H\).

If \(G\) acts on a set \(X\) and \(x, y \in X\), then \(x\) is said to be \(G\)-equivalent to \(y\) if there exists a \(g \in G\) such that \(gx =y\). We write \(x \sim_G y\) or \(x \sim y\) if two elements are \(G\)-equivalent.

Condensed: the full section is in Judson, Abstract Algebra: Theory and Applications.

Nou jy Geen sakrekenaar los hierdie een nie, maar die stukke daarvan is kon nie wees kondifiseerbaar. Probeer een hieronder, of tik jou eie.

Hou jou eie werk

'n Vry rekening voeg notas by oor elke les, 'n verslag van wat jy klaar gemaak het, jou opgelos probleme op een plek, en' n tugmeester wat jy kan vra oor hierdie bladsy. Die wiskunde self is oop vir almal, geteken in of nie.

Meld op Meld aan

Simbools wat hier gebruik word

Maak enige simbool van die volledige definisie,'n prent en wat elke letter daarin beteken.

Vrae wat mense vra

What is a group, in plain words?

A set with one operation that is associative, has an identity, and lets every element be undone. Symmetries of any object form a group. That is where the idea came from.

What is the difference between a ring and a field?

A ring has addition and multiplication that behave like the integers (you cannot always divide); a field is a ring where every non-zero element has a reciprocal, like the rationals or the reals.

Dele van hierdie bladsy word aangepas Judson, Abstract Algebra: Theory and Applications (GFDL 1.3). Verklaar en weer opklaar hier, foute is ons s'n.

Meer in Abstract Algebra

Groups Subgroups, cosets and Lagrange's theorem Cyclic groups and permutation groups Homomorphisms, normal subgroups and quotient groups Rings and fields Galois theory: why the quintic has no formula