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Flashcards

Math Proofs

amsond01 Inglês 28 flashcards

Flashcards do conjunto (28)

  1. 1Prove that if G is a group, G has a unique inverseSuppose b,c are both inverses of a. Then ab=e and ac=e so that ab=ac. Usiing left cancellation, b=c.
  2. 2Suppose G is a group, and a,b are element of G. with a an element of H but b not an element of H prove that ab is not an element of Ha and b are elements of G, therefore a-1 and b-1 are elements of G. Suppose ab element of H it implies ab-1 is an element of H. If H is a subgroup it is closed under the operation and b-1 is an element of H and therefore b is an element of H.
  3. 3Prove that is a^2b^2 =(ab)^2 than a and b commuteFor a2b2=(ab)2 it follows aabb=abab. Using left cancellation, it follows abb=bab. Using right cancellation it follows ab=ba.
  4. 4Let L:G-G be a function given by L(u)=au for all u element of G. Prove that L is one to oneFor L to be one to one must show for L(x) and L(y) x=y for x,y element of G. Therefore L(x)=ax
    L(y)=ay, so ax=ay, using left cancellation x=y.
  5. 5Prove that a^-1b element of H implies that aH=bH.If aH=bH, then a=ae element of aH=bH. Conversely, if a element of bH we have a=bh where h element of H and therefore aH=(bh)H=b(hH)=bH.
  6. 6Prove that the function L:H-bH for all h element of H is a bijectionFor onto must show L(x)=y where x is an element of bH and x is an element of H and L(x)=y let x=b-1y
    L(b-1y)=bb-1y=y. For one to one, let x,y element of H and L(x)=L(y) therefore, bx=by. Using left cancellation x=y.
  7. 7Prove in G there is only one identity elementSuppose e and e' are the identity, then ae=a and ae'=a. Therefore ee'=e' and ee'=e this is only true if they are equal to each other.
  8. 8Suppose b E Z and a E N and that q1,q2,r1,r2 E Z with a>r1>0 and a>r2>0 and b=aq1+r1=aq2+r2 Prove r1=r2 and q1=q2For aq1+r1=aq2+r2 it follows, aq1-aq2=r2-r1 and it follows, a(q1-q2)=r2-r1. It is shown a divides r2-r1. For r2>r1, it follows, a>r2-r1>0 so r2-r1=0, and r2=r1 and q1=q2.
  9. 9Suppose ab E Z-(0) and d=gcd(a,b). Use the Rel Prime Corollary to prove gcd(a/d,b/d) = 1For d= gcd(a,b) it follows a/d=x and b/d=y are relatively prime. From Corollary, if a and b are relatively prime there exists a gcd(x,y) such that gcd(x,y)=1.
  10. 10Suppose G is a group, and b E G and |b|=4. Use the division theorem to prove that =(e=b0,b1,b2,b3).All elements are contained in . Let u E such that u=b^n, where n E Z, the order of u divides the order of b. Therefore, n=4q+r where q,r E Z such that 0<=r<4. Using the division theorem b^n=b^4q+r=(bq)^4b^r Since b^4 is the identity b^n=b^r and 0<=r<4 so u must be an element of (b0,b1,b2,b3)
  11. 11Let G be a group and let A and B be subgroups of G. Prove that A intersection of B is a subgroup.Using one step subgroup test. Let x,y E A intersection B and x,y E A. This implies xy-1 E A. For x,y Eb implies xy-1 E B. Therefore xy-1 is an element of A intersection of B.
  12. 12Prove if G is abelian and N is a normal subgroup of G than G/N is abelian.Let g1N and g2N be arbitrary elements of G/N. Because N is normal it follows g1Ng2N=g1g2N. Since G is Abelian, g1g2N=g2g1N. Since N is normal it follows g2g1N=g2Ng1N. Using arbitrary elements in G/N is is shown g1Ng2N=g2Ng1N.
  13. 13If G is cyclic with generator b, and N is a normal subgroup of G, then G/N is a cyclic generator bN.Since b is the generator of G =G consider that =G/N. If Ng E G/N than g E G and g=b^m for m E Z. Therefore (Nb)^m = N^mb^m=Nb^m=Ng. We conclude that G/N subgroup so that G/N= and therefore cyclic.
  14. 14If G is a group whose order is a prime number, then G is cyclic.The order of an element is the order of the subgroup generated by that element. Suppose G has prime. Let a E G and a is not the identity. Then, || divides |G| and || is not 1, therefore ||=|G|
  15. 15If G is a group Prove ab=ac then b=cIf G is a group and a,b,c E G and ab=ac than the group is closed under the operation and using left cancellation b=c.
  16. 16Prove that if y mod n =1 than gcd(y,n)=1By Rel. Prime Corollary, let a,b E Z-(0) then they are rel. prime iff there exists s,t E Z such that as+bt=1. Suppose n E N and aEZ and a mod n = 1 therefore there exists a=nq+1 with (-1,q) in the role of (s,t) above the Rel. Prime Corollary guarantees that a and n are Rel. Prime
  17. 17Let G be a group, bEG and |b|=nEN. Let cEG. Prove that |cb-1c|=|b|=nLet c,b E G and let x=cbc-1 the order of b is n, therefore b^n=e.
    x^n=(cbc-1)(cbc-1)cbc-1)...n times
    thus, x^n=b^n=e. Conversely, b=c-1xc and b^n=(c-1xc)(c-1xc)...n times and is also equal to the identity.
  18. 18Let G be a group, bEG and |b|=nEN. From the Division theorem, prove that for all mEZ, b^m=e iff n|m.Let a^m be a member of . By the division theorem there exists integers r,q such that m=nq+r such that 0<=r
  19. 19List All S3 subgroupse, S3, Z3, S2, Z2, A3
  20. 20Once Proven =(e,b1,b2,b3> then prove ||=4The definition of order is the number of elements of the group. There are four elements in therefore the order is four.
  21. 21Prove bHb-1 is a subgroup of GUsing one step subgroup test, let x,y E bHb-1. Will show xy-1E bHb-1. Let h,h2 E H For x=bhb-1 and y=bh2b-1 Let h E H and xy-1 = (bhb-1)(bh2b-1)=bhh2b-1. Since H is a subgroup and hh2 E H and it follows bhh2h-1 E bHb-1 and it is a subgroup or G.
  22. 22Prove bHb-1 is isomorphic to H, under the map a:H-bHb-1 given by a(h)=bhb-1 for all hEHMust show that it is a bi-jection and it preserves the function (1)ONTO: Let y E bHb-1 and x E H for a(x)=bxb-1 (2) One-to-One: Let x,y E H and a(x)=bxb-1 and a(y)=byb-1 Thus, bxb-1=byb-1. Using left and right cancellation x=y. (3) Preserves the function let x,y E H for a(xy)=bxyb-1=bxb-1byb-1=a(x)a(y)
  23. 23Prove that ker(x) is a normal subgroup of G.Let a,b E ker(x) need to show ab-1Eker(x) to prove it is a subgroup. Since x(a)=e and x(b)=e we have x(ab-1)=x(a)x(b-1)=x(a)x(b)-1=ee-1=e therefore ab-1E ker(x). Next to show normal, need to prove for gEG gker(x)g-1 E ker(x). let k E ker(x).
    x(gkg-1)=x(g)x(k)x(g-1)=x(g)x(g-1)=eg and thus gkg-1 E ker(x)
  24. 24Prove H/N is a subgroup of G/NUse one step subgroup test, let h1N, h2N E H/N, assuming h1,h2 E H must prove h1N(h2N)-1 E H/N. h1N(h2N)-1=h1N(h2-1)N=h1h2-1N and since H is a subgroup closec under inverse h1h2-1 is an element of H/N
  25. 25Prove that b,c are in G, then $(b)=$(c) IFF bker($)=cker($)Assume $(b)=$(c), then e=($(b))-1$(c)=$(b-1)$(c)=$(b-1c) so that b-1cEker($). It follows that bKer($)=cKer($). Reverse the argument to complete the proof.
  26. 26Prove $(e)=eSince ee=e, $(e)=$(ee)=$(e)$(e) and since $(e) is an element of G, e$(e)=$(e) by cancellation $(e)=e
  27. 27Prove $(a-1)=($(a))-1Let, g E G. For, e=$(e)=$(gg-1)=$(g)$(g-1)=$(g)($(g))-1
  28. 28Prove that b E G and |b|=nEN than the order of $(b) divides |b|For the order of b, b^n=e. This implies e=$(e)=($(b^n))=$(b)^n. So by corollary a^n=e than the order of a divides n.

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Descrição
FlashCards sobre Math Proofs, criado por amsond01 em 08-12-2015.
Criado por
amsond01 · dez 2015
Atualizado
27 abr 2016
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