Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Problem regarding computation of factor groups.

  1. Oct 10, 2009 #1

    I am quite new here, as my number of posts might indicate. Thus I am not really sure whether or not this question should be posted here or somewhere else. It is not a homework, but neither is it a question that could not be a homework. However, here we go.

    I have, during a course in Abstract algebra, solved problems of the type:

    Compute, i.e. classify according to the Decomposition Theorem (DT) for finitely generated abelian groups, The factor group (Z*Z*Z) / <(3,3,3)>.

    The solution to this problem is not very hard. One just constructs a homomorphism f: Z*Z*Z --> Z*Z*Z having kernel <(3,3,3)>, and then uses the theorem (sometimes called the fundamental homomorphism theorem, sometimes the first isomorphism theorem) that states that if f:G --> G' is a homomorphism with ker(f) = H, Then G/H is isomorphic to Im(f). In the case above, Im(f) would be (Z/3Z)*Z*Z, or some isomorphic group that is later translated to the appropriate form.

    The normal subgroup <(3,3,3)> can be alternated to make this problem more cumbersome.

    One case I have not yet managed to complete is when <(3,3,3)> is replaced by H = <(6,4,8)>. Here I seem to fail when trying to restrict my homomorphism to some subset of the original codomain.

    Now, if the problem had been to compute for example (Z*Z*Z*Z*Z*Z) / <(6,4,8,5,12)> I would certainly give up before even trying. Constructing a homomorphism with the appropriate kernel is perhaps not too hard, but finding the image... I would be not too cheerful at the thought.

    Let's, for simplicity of notation, let G = Z*Z*...*Z, H = <(a1, a2, ..., an)> and Let K be the group isomorphic to G/H expressed as a product of cyclic groups according to DT.

    My question is, has anyone got any ideas on how to simplify the work here? Could conclude something about the Betti number of K just be regarding the dimension of G and the integers a_i, (i = 1,2,...,n)? If the ai are all equal, or such that they are all multiples of one a_j, then one can just construct easily homomorphism reducing the a_j'th coordinate of G (mod aj), and then manipulate the other coordinated by some linear combination such that the appropriate kernel is obtained, and it is easy to find a proper restriction of the homomorphism, so that one can apply the fundamental homomorphism property.

    I would appreciate hints and tips from more knowledgable people, as I would like to gain some understanding of this type of quotient groups, and it would simplify computations.

    Thanks in advance
  2. jcsd
  3. Oct 10, 2009 #2


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    This is a form of linear algebra.

    Your first example of Z³ / <3,3,3> is simplified by doing column operations to reduce the vector
    to the vector
    (Note that column operations have to have integer coefficients and invertible over the integers -- meaning, e.g., that we can't multiply or divide by 3)

    This works because it's just a change of basis. If E is the elementary matrix used to enact the column operations you did, then we use E to define an isomorphism f:Z³ --> Z³.

    If H is a subgroup of the abelian group G, and f:G --> G' is an isomorphism, can you prove the following?
    [tex]G / H \cong G' / f(H)[/tex]​
    (Hint: it's easy to give an explicit definition of the isomorphism....)
  4. Oct 10, 2009 #3
    I guess if f: G --> G' is an isomorphism of abelian groups, H is a subgroup of G it follows directly (from some correspondence thm for groups) that f(H) is a subgroup of G'. As they are abelian and thus normal, we can consider the quotient group G'/f(H). Define now the map h: G/H --> G'/f(H) by h(gH) = f(g)f(H).

    h is clearily surjective, as if af(H) is an element of G'/f(H), then there exists a unique a' in G such that f(a') = a. Thus h(a'H) = f(a')f(H) = af(H).

    h is also injective, as if gH is not equal to kH, then there exists no x in H such that g = kx. But then, if we assume that h(g) = h(k), that is: f(g)f(H) = f(k)f(H), then there exists some y in f(H) such that f(g) = f(k)y. But as f is an isomorphism, there exists an x in H such that f(x) = y. This gives f(k)f(x) = f(h), which by the homomorphism property of f is equivalent with f(g) = f(kx), and by injectivity of f, also g = kx, which is a contradiction.

    The homomorphism property:By the fact that f is an isomorphism and by the definition of multiplication in factor groups; h(gHkH) = h(gkH) = f(gk)f(H) = f(g)f(k)f(H) = f(g)f(H)f(k)f(H) = h(gH) h(kH).

    Thank you for your hint. Though I must say, as I haven't read any linear algebra (still in my first semester), that I am just as lost when the normal subgroup gets more complicated as before. Though, following your hint, I think we could reduce it to for example Z^3 / <(2,0,0)>, ([2, 0, 0] obtained by subtracting 4 from 6, then subtracting 2*4 from 8 and last subtracting 2 from 4), which would then collapse the first factor (mod 2), leaving the rest untouched, giving us an isomorphism of Z^3 / <(6, 4, 8)> with (Z/2Z)* Z^2. Though, this feels not so good, for some reason.

    By the way, two more questions: How do I turn to TeX-mode? and :Are the restrictions on f (the column operations) necessary to ensure us that G' (in the notation of the isomorphism result above) is invariant under f?

    Edit: Having read about the necessary notions in the field of linear linear algebra, I think I understand what was done here, more precisely. Though, a comment on the solution of the special case would be highly appreciated.
    Last edited: Oct 10, 2009
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Similar Discussions: Problem regarding computation of factor groups.
  1. Factor Group (Replies: 5)

  2. Factor Groups (Replies: 1)

  3. Factor groups (Replies: 2)