Problem regarding computation of factor groups.

1. Oct 10, 2009

Jösus

Hello,

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.

2. Oct 10, 2009

Hurkyl

Staff Emeritus
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
[3,3,3]​
to the vector
[3,0,0].​
(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?
$$G / H \cong G' / f(H)$$​
(Hint: it's easy to give an explicit definition of the isomorphism....)

3. Oct 10, 2009

Jösus

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