MTH 4- Astrct Alger II S7 Solutions to Homework 6 Exercise () Let A, B n C e groups with A B n B C Show tht A C () (This ws not prt of Exercise, ut shoul hve een) Let A n B e groups with A B Then B A () Since A B n B C, there exist isomorphism f A B n g B C By Exercise 7 on Homework 3 g f A C is homomorphism Since f n g re isomorphism, they re ijections Hence y Lemm A23(c), g f is ijection Thus g f is n isomorphism from A to C Hence A C () Since A B, there exist n isomorphism f A B Then f is ijection n so y Lemm A26(c) f hs n inverse, tht is function g B A with f g = i B n g = i A By Lemm A26(,), we know tht g is - n onto, so g is ijection We will now show tht g is homomorphism Note tht for ny B we hve ( ) f(g()) = (f g)() = i B () = Let, B n put ( ) = g() n = g( ) By ( ) pplie to n we hve ( ) f() = n f( ) = We compute f( ) =f()f( ) = f is homomorphism ( ) =f(g( )) ( ) pplie to We prove tht f( ) = f(g( )) Since f is - we infer g( ) = ( ) = g()g( ) Thus g is homomorphism We lrey prove tht g is ijection, so g is n isomorphism Hence B A Exercise 2 Let A n B e groups with A B Show tht A is elin if n only if B is elin Since A B there exists n isomorphism φ A B : Suppose first tht A is elin Let, B Since φ is n isomorphism, φ is onto So there exist, c A with ( ) φ() = n φ(c) =
Since φ is isomorphism, φ is homomorphism We hve = φ()φ(c) ( ) = φ(c) φ is homomorphism = φ(c) A is elin = φ(c)φ() φ is homomorphism = ( ) Thus = for ll, B n B is elin : Suppose next tht B is elin I will give two proofs tht A is elin Proof : By Exercise () we know tht B A Hence y the lrey proven forwr irection of this exercise (pplie with B n A interchnge) we conclue tht A is elin Proof 2: Let, c A Then φ(c) = φ()φ(c) φ is homomorphism = φ(c)φ() B is elin = φ(c) φ is homomorphism We prove tht φ(c) = φ(c) Since φ is n isomorphism, we know tht φ is - So c = c for ll, c A n A is elin Exercise 3 Let φ A B e n isomorphism of groups Let C A n put D = φ(c) Show tht () C D () D B (c) A/C B/D () Define φ C C B, c φ(c) By 93(c), φ C is n homomorphism n since φ is -, so is φ C Hence y 65, Im φ C is sugroup of B isomorphic to C We hve n so () hols Im φ C = {φ C (c) c C} = {φ(c) c C} = φ(c) = D () Since φ is n onto homomorphism, 96 () shows tht φ(c) B Since D = φ(c), () hols (c) Let π B B/D, D e the nturl homomorphism By 94 π is n onto homomorphism Put α = π φ By Homework 3#7 α is homomorphism Since π n φ re onto, α is onto, see Lemm A23() Thus 2
( ) Im α = B/D Let A Then Thus Kerα α() = e B/D efinition of Kerα (π φ)() = e B/D efinition of α π(φ()) = e B/D efinition of φ() Kerπ efinition of Kerπ φ() D 94 φ() φ(c) efinition of D φ() = φ(c) for some c C efinition of φ(c) = c for some c C φ is - C ( ) Kerα = C By the First Isomorphism Theorem n so y ( ) n ( ): A/Kerα Im α A/C B/D Exercise 4 Let G e group, N G n H G Suppose tht G/N (Z 24, +), tht H = 8 n tht N is o Show tht H N = {e} n H (Z 8, +) Oserve tht the function i H H H, h h is homomorphism with Ker i H = {e} n Im i H = H Hence the First isomorphism Theorem shows tht ( ) H H/{e} Since H is power of 2 n N is o we hve gc ( H, N ) = Thus Exercise 3 on Homework 4 shows tht H N = {e} By the Secon Isomorphism Theorem we hve H/H N HN/N Thus ( ) H/{e} HN/N By the Corresponence Theorem HN/N is sugroup of G/N Since G/N Z 24 n HN/N G/N, we conclue from Exercise 3() tht 3
( ) HN/N K for some K Z 24 By 99(c) (+) K Z l for some integer l From ( ), ( ), ( ), (+) n Exercise we hve H Z l In prticulr, l = Z l = H = 8 n so H Z 8 Exercise 5 Let G e group, N norml sugroup of G n efine Show tht is well-efine ction of G on N G N N, (g, n) gng Since N G, Lemm 86(f) shows tht gng N for ll g G n n N Thus is well-efine Let, G n n N e n = ege = ege = n n so (ct i) hols ( n) = (n ) = ()n( ) 48 = ()n() = () n So lso (ct ii) hols n is well efine ction Exercise 6 Let G e group cting on the set I Let J e suset of I Put Show tht St G (J) is sugroup of G Let g G By efinition of St G (J) we hve St G (J) = {g G gj = j for ll j J} ( ) g St G (J) if n only if gj = j for ll j J Let, St G (J) Then y ( ): ( ) j = j n j = j for ll j J We will now verify the three conitions of the Sugroup Proposition: (i) By (ct i) ej = j for ll j J n so ( ) gives (ii) for ll j J n so y ( ) ()j e St G (J) ct ii = (j) ( ) = j ( ) = j St G (J) 4
(iii) Let j J By ( ) j = j n so y 27(c) j = j Thus ( ) implies tht St G (J) We verifie the three conitions of the Sugroup Proposition n we conclue tht St G (J) is sugroup of G Exercise 7 Let G = GL 2 (R) Recll from Exmple 22(3) tht G cts on R 2 Compute St G (J) for J = {( 0 )}, J = {( )} n J = {( 0 ), ( )} Let A = [ c ] G Suppose first tht J = {( 0 )} Then A St G(J) if n only if if n only if n if n only if c =, 0 0 = c 0 = n c = 0 Note tht [ 0 ] is invertile if n only if 0 Hence St G (J) = R, R 0 Suppose next J = {( )} Then A St G(J) if n only if if n only if n if n only if Oserve tht c So [ ] is invertile if n only if Hence =, + = c + = n c = et = ( ) ( ) = + = 5
St G (J) =, R, Suppose finlly tht J = {( 0 ), ( )} Then A St G(J) if n only if c 0 = 0 n c = From the clcultion in the previous two cses this hols if n only if =, c = 0, =, c = n so if n only if =, = 0, c = 0 n = 0, tht is if n only if A = 0 0 Hence St G (J) = 0 0 6