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

  1. \(ex = x\) for all \(x \in X\);

  2. \((g_1 g_2)x = g_1(g_2 x)\) 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.

Sekarang Anda Tidak ada kalkulator yang bisa menuntaskan soal ini, tetapi bagian-bagiannya dapat dihitung. Coba salah satu di bawah ini, atau ketik soal Anda sendiri.

Simpan pekerjaan Anda sendiri

Sebuah akun gratis menambahkan catatan pada setiap pelajaran, catatan apa yang telah Anda selesaikan, masalah yang Anda selesaikan di satu tempat, dan tutor yang dapat Anda tanyakan tentang halaman ini. matematika itu sendiri terbuka untuk semua orang, ditandatangani atau tidak.

Daftar Masuk

Simbol yang digunakan di sini

Sentuh simbol apapun untuk definisi lengkap, gambar, dan apa setiap huruf di dalamnya berarti.

Pertanyaan yang sering diajukan

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.

Bagian dari halaman ini diadaptasi dari Judson, Abstract Algebra: Theory and Applications (GFDL 1.3). Diringkas dan dijelaskan ulang di sini; kesalahan adalah tanggung jawab kami.

Lainnya di Abstract Algebra