The inverse element is preserved, i.e. for any element (g, a) in the set, there exists an inverse element (g−1, a−1) such that (g, a) (g−1, a−1) = (1, 1) for the matrices.
To show that the following maps define a group action, we need to prove that the elements in the set are homomorphisms, i.e. that the action of a group element can be defined by multiplying the original element by another element in the group (by means of multiplication) for the matrices.
Let's examine each of the given sets in detail:(1) GL(n, R) × Matn(R) - Matn(R) defined as (A, X) → XA−1:To prove that this map defines a group action, we need to verify that the following properties are satisfied:The action is well-defined, i.e. given any two pairs (A, X) and (B, Y) in the set, we can show that (B, Y) (A, X) = (BA, YX) ∈ Matn(R). The identity element is preserved, i.e. given a matrix X ∈ Matn(R), the element (I, X) will be mapped to X.
The action is associative, i.e. given a matrix X ∈ Matn(R) and group elements A, B, C ∈ GL(n, R), the following equality will hold: [(A, X) (B, X)] (C, X) = (A, X) [(B, X) (C, X)]. The inverse element is preserved, i.e. for any element (A, X) in the set, there exists an inverse element (A−1, XA−1) such that (A, X) (A−1, XA−1) = (I, X).(2) (GL(n, R) × GL(n, R)) × Matr(R) -› Matn(R) defined as ((A, B), X) → AXB−1:Let's again verify the following properties for this map to define a group action: The action is well-defined, i.e. given any two pairs ((A, B), X) and ((C, D), Y), we can show that ((C, D), Y) ((A, B), X) = ((C, D) (A, B), YX) ∈ Matn(R). The identity element is preserved, i.e. given a matrix X ∈ Matn(R), the element ((I, I), X) will be mapped to X. The action is associative, i.e. given a matrix X ∈ Matn(R) and group elements (A, B), (C, D), E ∈ GL(n, R), the following equality will hold: [((A, B), X) ((C, D), Y)] ((E, F), Z) = ((A, B), X) [((C, D), Y) ((E, F), Z)].
The inverse element is preserved, i.e. for any element ((A, B), X) in the set, there exists an inverse element ((A−1, B−1), AXB−1) such that ((A, B), X) ((A−1, B−1), AXB−1) = ((I, I), X).(3) R × R2 → R2 defined as (r, (x, y)) → (x + r4, y):Again, let's check the following properties to show that this map defines a group action: The action is well-defined, i.e. given any two pairs (r, (x, y)) and (s, (u, v)), we can show that (s, (u, v)) (r, (x, y)) = (s + r, (u + x4, v + y)) ∈ R2.
The identity element is preserved, i.e. given an element (x, y) ∈ R2, the element (0, (x, y)) will be mapped to (x, y). The action is associative, i.e. given an element (x, y) ∈ R2 and group elements r, s, t ∈ R, the following equality will hold: [(r, (x, y)) (s, (x, y))] (t, (x, y)) = (r, (x, y)) [(s, (x, y)) (t, (x, y))]. The inverse element is preserved, i.e. for any element (r, (x, y)) in the set, there exists an inverse element (-r, (-x4, -y)) such that (r, (x, y)) (-r, (-x4, -y)) = (0, (x, y)).(4) FX × F → F defined as (g, a) → ga, where F is a field, and FX = (F \ {0},) is the multiplicative group of nonzero elements in F:To show that this map defines a group action, we need to verify that the following properties are satisfied:The action is well-defined, i.e. given any two pairs (g, a) and (h, b), we can show that (g, a) (h, b) = (gh, ab) ∈ F.
The identity element is preserved, i.e. given an element a ∈ F, the element (1, a) will be mapped to a. The action is associative, i.e. given elements a, b, c ∈ F and group elements g, h, k ∈ FX, the following equality will hold: [(g, a) (h, b)] (k, c) = (g, a) [(h, b) (k, c)]. The inverse element is preserved, i.e. for any element (g, a) in the set, there exists an inverse element (g−1, a−1) such that (g, a) (g−1, a−1) = (1, 1).
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