proof of fourth isomorphism theorem

First we must prove that the map defined by AA/N is a bijection. Let θ denote this map, so that θ(A)=A/N. Suppose A/N=B/N, then for any aA we have aN=bN for some bB, and so b-1aNB. Hence AB, and similarly BA, so A=B and θ is injectivePlanetmathPlanetmath. Now suppose S is a subgroupMathworldPlanetmathPlanetmath of G/N and ϕ:GG/N by ϕ(g)=gN. Then ϕ-1(S)={sG:sNS} is a subgroup of G containing N and θ(ϕ-1(S))={sN:sNS}=S, proving that θ is bijectiveMathworldPlanetmath.

Now we move to the given properties:

  1. 1.

    AB iff A/NB/N

    If AB then trivially A/NB/N, and the converseMathworldPlanetmath follows from the fact that θ is bijective.

  2. 2.

    AB implies |B:A|=|B/N:A/N|

    Let ψ map the cosets in B/A to the cosets in (B/N)/(A/N) by mapping the coset bA bB to the coset (bN)(A/N). Then ψ is well defined and injective because:

    b1A=b2A b1-1b2A

    Finally, ψ is surjective since b ranges over all of B in (bN)(A/N).

  3. 3.


    To show A,B/NA/N,B/N we need only show that if xA or xB then xNA/N,B/N. The other cases are dealt with using the fact that (xy)N=(xN)(yN). So suppose xA then clearly xNA/N,B/N because xNA/N. Similarly for xB. Similarly, to show A/N,B/NA,B/N we need only show that if xNA/N or xNB/N then xA,B. So suppose xNA/N, then xN=aN for some aA, giving a-1xNA and so xAA,B. Similarly for xNB/N.

  4. 4.


    Suppose xN(AB)/N, then xN=yN for some y(AB) and since N(AB), x(AB). Therefore xA and xB, and so xN(A/N)(B/N) meaning (AB)/N(A/N)(B/N). Now suppose xN(A/N)(B/N). Then xN=aN for some aA, giving a-1xNA and so xA. Similarly xB, therefore xN(AB)/N and (A/N)(B/N)(AB)/N.

  5. 5.

    AG iff (A/N)(G/N)

    Suppose AG. Then for any gG we have (gN)(A/N)(gN)-1=(gAg-1)/N=A/N and so (A/N)(G/N).
    Conversely suppose (A/N)(G/N). Consider σ:g(gN)(A/N), the compositionMathworldPlanetmathPlanetmath of the map from G onto G/N and the map from G/N onto (G/N)/(A/N). gkerπ iff (gN)(A/N)=(A/N) which occurs iff gNA/N therefore gN=aN for some aA. However N is contained in A, so this statement is equivalnet to saying gA. So A is the kernel of a homomorphismPlanetmathPlanetmathPlanetmathPlanetmathPlanetmathPlanetmathPlanetmath, hence is a normal subgroupMathworldPlanetmath of G.

Title proof of fourth isomorphism theorem
Canonical name ProofOfFourthIsomorphismTheorem
Date of creation 2013-03-22 14:17:38
Last modified on 2013-03-22 14:17:38
Owner aoh45 (5079)
Last modified by aoh45 (5079)
Numerical id 9
Author aoh45 (5079)
Entry type Proof
Classification msc 20A05