Question

a). Describe a natural group action of G on X. b). Show that the action has the required properties. (0) G= {e,a}, the cyclic group of order 2, X the vertices of an equilateral triangle. (ii) G=22 x Z2, X the vertices of a square. • 14.4: 2, 3, 5, 6(ac), 7.
a and b please. abstract algebra question

Answer 1

Font we give the defination of group action A group action of a group G on a Set A is a map Gtx A LOA written an gia gt, vatA satisfying : 2:(0-0) = (:)(a) v 3,€ GIQ6A 1.a = a vata where t is identity of G. . l. Q = 0 Va ♡ ut X= vertices of equilateral triangle ABC - G= {e,a} be cyclic group of order 2 - w e denotes "doing nothing" a denotes " fliping about the axis AA" Then . er=2 VREX • e.(Q.A)-CA - A = (e.a). A = a.le.A) e. (Q.B) -e.coC (ea)-B = a.(e.B) e. (a.) - 0 = B = (e.a).oc -ale.c) So, it is a gouocp action of G on X (ii) ut x = verotices of square ABCD AF 61 = 72x72- (0,0),(0,1),(1,0), Gis a commutative group. et cool denotes do nothing » (0,1) denotus "190° clock wise sulation" (10) denotes “ Flip about diagonal AC" Then (11) denotes a finst flip about diagonal Ac then Jotate 180° clock wise (0,0).x = x W xex (0,0)(0,1). A)= (0,0).C=C =((0,0).(0,1)). A = (0,1){(0,0).A) (0,0) (0,0). A) = (0,0). A-A-((0,0).(1,0)) A = (1,0). ((0,0). A) (0,0). (11,7. A) = (0,0).C=C =((0,0). (1,1). A = (1,1).((0,0). A) Then (5,1). (10.1). A) - (0,1).C= A - (691).(0,1)). A = (0,1).(0,1). A) (0-1){(1,0). A) = (0,1) A = 6 = (101)0,1)). A = (1,0).(10.19. AY

(0,1). (0,1). A) = (0,1).C = A = ((0,1).(1,1)). A = (1,1).(0,1).A). (1,0). ((1,0). A) = (1,0). A = A = 1 (1,0).(1,0)). A = (1,0). ((,0).A) (1,0).((1,1). A) = (0,0).C - C = ((1,0),(1,1)). A = (1/1). (C10). A) So, 9, (824) = 18.92)A • & g, .92 € 72 x 22 Similarly we can show that 9, (92:B) = (9,82) B & 8, 92€ 72 4742 9,. (92. c) = 18,092).C & g. 92 E 2 X 72 9,. (92.0) = 18, 92). & 8., 92f P2 X 722 It is a group action of G on x Hence

Answer 2

Solution :

(a)

(i)

Suppose that the vertices of the equilateral triangle be labelled 1,2,3. In other words, let X={1,2,3}.

Define the action . : G x X $\rightarrow$ X given by, e.n = n for all n=1,2,3 and a.1 = 1, a.2 = 3, a.3 = 2. In other words, the action of the group element a on X is to reflect the equilateral triangle about the median passing through 1.

(ii)

Let the vertices of the square be labelled 1,2,3,4.

Equivalently, let X = {1,2,3,4}

Define the action . : G x X $\rightarrow$ X by,

(0,0).n = n for all n=1,2,3,4

(1,0).1 = 2, (1,0).2 = 1, (1,0).3 = 4, (1,0).4 = 3

(0,1).1 = 3, (0,1).3 = 1, (0,1).2 = 4, (0,1).4 = 2

(1,1).1 = 4, (1,1).2 = 3, (1,1).3 = 2, (1,1).4 = 1

In other words, the action of (1,0) is to reflect the square along the central line perpendicular to the segments [1 2] and [3 4]. The action of (0,1) is to reflect the square along the central line perpendicular to the segments [1 3] and [2 4]. The action of (1,1) is the composite of the previous two actions.

(b)

The fact that the two maps defined above are indeed group actions of G on X is evident.

Firstly, by definition, in each case, e.a = a for all a in X. Secondly, g₁.(g₂.a) = (g₁g₂).a for all g₁,g₂ in G and a in X.

Hence, . is indeed a group action in the above two cases.