maths.freeAbstract Algebra › 14. 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.

Symbols used here

\lambda
lambda (eigenvalue)
The factor by which an eigenvector is stretched: Av = λv.
x \in A,\ A \subseteq B
element of, subset
x belongs to A; every element of A is in B.
\sigma,\ s,\ \sigma^2
standard deviation, sample s.d., variance
Typical distance from the mean; its square.
\mathbb{N},\ \mathbb{Z},\ \mathbb{Q},\ \mathbb{R},\ \mathbb{C}
number sets
Naturals, integers, rationals, reals, complex numbers.
\blacksquare\ \text{or}\ \square
end of proof (halmos)
Marks the point where the statement has been established.
a \equiv b \pmod n
congruent modulo n
n divides a − b; a and b have the same remainder.
a \mid b,\ \gcd(a,b)
divides, greatest common divisor
b is a multiple of a; the largest number dividing both.
(G, \cdot),\ e,\ g^{-1}
group, identity, inverse
A set with an operation; the do-nothing element; the element that undoes g.
G \cong H,\ G / N
isomorphic, quotient group
Same structure; the group of cosets of a normal subgroup N.
\mathbb{Z}/n\mathbb{Z},\ \mathbb{Z}_n
integers modulo n
The remainders 0…n−1 with clock arithmetic.
\operatorname{Hom}(A, B),\ f \circ g
arrows from A to B, composition
The set of morphisms; do g then f.

Questions people ask

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.

Subukan ang iyong sarili

Parts of this page are adapted from Judson, Abstract Algebra: Theory and Applications (GFDL 1.3). Condensed and re-explained here; errors are ours.

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