ANVANDED ABSTRACT ALGEBRA

ANVANDED ABSTRACT ALGEBRA ACTIONS OF GROUP ON A SET Let G be a group Z(G) = { z G | zx=xz x G} EX1, Check that Z(G) is a subgroup of G i) x G, one has : 1Gx=x1G ( 1G G)   ii)  1G Z(G) Z(G) z1,z2 Z(G) z1x=xz1; z2x=xz2 (*) x G) x G, z1,z2 Z(G), one has: (z1z2)x = z1( z2x) = z1(xz2) = (z1x)z2 = (xz1)z2 = x(z1z2)  iii)    z1z2 Z(G) (**) x G => x-1 G ( since G be a group) z Z(G), x G, one has: zx-1 = x-1z (zx-1)-1 = (x-1z)-1 xz-1 = z-1x z-1 Z(G) (***) From (*) and (**) and (***) , this implies Z(G) is subgroup of G EX2,Let G ba a group such that G/Z(G) is cyclic Prove that G = Z(G) if G is abelian •  •    G/Z(G) is cyclic group g G such that G/Z(G) = g1,g2 G g1Z(G), g2Z(G) m,n N such that g1Z(G) = gmZ(G), g2Z(G) = gnZ(G) g1 = gmk, g2 = gnk’ ( k, k’ Z(G)) g1,g2 G,one has: g1 g2 = (gmk)(gnk’) = gm (kgn )k’ = gm (gnk )k’= gm+n( k k’) g2 g1 = (gnk’)(gmk) = gn (k’gm )k = gn (gmk’ )k= gm+n( k’ k) Since k,k’ Z(G) => kk’ = k’k So, g1 g2 = g2 g1  •    G is abelian We see that : Z(G) G Since G is Abelian so g G gx = xg x G g Z(G) G Thus G = Z(G) if G is Abelian EX3, Let p be a prime integer Prove that any group G with p2 element is Abelian G be a finite group with p2 element ( p be a prime number) Then Z(G) has more than one element so by Lagrange’s theorem |Z(G)| is a divisor of |G| On the other hand , p be a prime integer thus Z(G) has oder p2 or p •   • If Z(G) has oder p |G/Z(G)| = G/Z(G) has order p On the other hand, p is prime number thus G/Z(G) is cyclic By EX2 G is Abelian so G = Z(G) ( contracdiction since |G| ) So this case cannot arise If Z(G) has oder p2 We see that Z(G) G On the other hand | Z(G)| = |G| = p2 < + Thus Z(G) = G Since Z(G) is Abelian so G is Abelian EX4) A group with p2 elements is either cyclic or isomorphic to the direct product • If G has an element a of order p2, then G is cyclic group G = •  If G has no element of oder p2 Every non-identity elements of G has oder p Let x be any such element, so that has p elements Now we choose any element of G not in Since = {1} It follows that the p2 elements {xixj: are all disctinct and so are all the elements of G it may then be checked that the map f : xixj (xi,xj) One has f (xixj) = Is, in fact, an isomorphism between G and direct product of the groups We conclude thus section with a generalization of a fact we have already established http://faculty.mu.edu.sa/public/uploads/1336339374.2812groups_t heory.pdf http://www.abstractalgebra.net/Lectures/L23.pdf http://www.math.binghamton.edu/mazur/teach/40107h42sol.pdf http://www.willamette.edu/~emcnicho/courses/AbstractAlgebra/Ex amsF09/GExam1Soln.pdf http://faculty.mu.edu.sa/public/uploads/1336339374.2812groups_theory.pdf Xác định lớp liên hợp http://www.math.uconn.edu/~kconrad/blurbs/grouptheory/conjclass.pdf A4 nhóm phép chẵn EX5: Let G be the symmetric group S(n) and V be a complex vector space with basis { } For G and and of V, define • i) n = and ii) n = and Show that V is a G- set - One has = = =) =  Thus V is a G-set i) Since  = S(4) Orb() = = } = ii) Since  = { 1, (2 4),(1 3),(1 3)(2 4)} Orb() = = } ={ EX6: Let X be a G-set and Show that for any g G, the stabiliser of g; is the subgroup The stabiliser of g is the set { ={ ={ ={ = EX7: Determine the list of conjugacy classes in the symmetric group S(4) and also in the alternating group A(4) • Conjugacy classes in S(4) are: {1} { (1 2), (1 3),(1 4),(2 3),(2 4),(3 4)} { (1 3), (1 2), (1 4),(1 3),(1 4),(1 2), ( 4),(2 3)} {( 4),(1 4),(1 2),(1 3),(1 2)} { (1 2)(3 4), (1 3)(2 4), (1 4)(2 3)} • Conjugacy classes in A(4) {1} {( 2), (2 4),(1 3), (1 4)} {(1 3), (1 4), (1 2),( } { (1 2)(3 4), (1 3)(2 4), (1 4)(2 3)} EX8: Let G be any group and g be an element of G Prove directly that is a subgroup of G • One has   (*) • x1,x2  x1g=gx1; x2g=gx2 x ) xg G, x1,x2 one has: (x1x2)g = x1( x2g) = x1(gx2) = (x1g)x2 = (gx1)x2 = g(x1x2)  •    x1x2 (**) g G => g-1 G ( since G be a group) x , g G, one has: xg-1 = g-1x (xg-1)-1 = (g-1x)-1 gx-1 = x-1g x-1 (***) From (*) and (**) and (***) , this implies is subgroup of G EX10 : Let G be the group S(3) Calculate when H is a) Subgroup {1, (1 3), (1 2)} b) Subgroup { 1,(1 2)} a) One has : g ( 2) (1 3) (2 3) (1 3) (1 2) 1 1 1 (1 3) (1 2) (1 2) (1 2) (1 3) (1 3) (1 2) (1 3) (1 3) (1 3) (1 2) (1 2) We see that Thus = G b) One has g ( 2) (1 3) (2 3) (1 3) (1 2) 1 1 1 (1 2) (1 2) (3 2) (1 3) (2 3) (3 1)  = {1, (1 2)} = H EX9: Let G In any group the identity element always forms a conjugacy class on its own, so if G has precisely two conjugacy classses, all the non-identity elements lie in one class Since the number of elements in a conjugacy class divides n = \G\, we deduce that n — divides n It then follows that n — divides n — (n — 1) = so that n — = and hence n =  ... heory.pdf http://www.abstractalgebra.net/Lectures/L23.pdf http://www.math.binghamton.edu/mazur/teach/40107h42sol.pdf http://www.willamette.edu/~emcnicho/courses/AbstractAlgebra/Ex amsF09/GExam1Soln.pdf