Let G,H,K be finitely generated abelian groups. Show that if G*H is isomorphic to G*K , then H is isomorphic to K, where * represents the external direct product of groups.
One simple idea of this problem is to divide G,H,K into Z(p1^k1)*Z(p2^k2)...,but the question is to prove the unity of such representation, that is to say, is it possible that Z(6)*Z(2) isomorphic to Z(2)*Z(2)*Z(3), or the more general situation?
This is actually true.
As far as proving this I don't think it is that complicated.
If , then there is an isomorphism so say
, but then just look at the projection onto the second coordinate of this map. It inherits surjectivity, homomorphism, welldefinedness, from the fact that is an isomorphism.
It will go from , it clearly has kernel G, and
we may write the isomorphism as F(g,h)=(f1(g,h),f2(g,h))
and we have f1(g1g2,h1h2)=f1(g1,h1)f1(g2,h2)
how to ge the kernel? that is, to solve the equaltion f2(g,h)=e??
do you mean f2(g,e)=e?but what i can deduce is f2(e,e)=e??
I am confused..
yeah, kernel needs to be , you are right, but this is pretty obviously isomorphic to just G by just dropping the e part. It can be confusing sometimes when you are dealing with something so obviously true I think. If you believe you can just write the isomorphism in terms of the coordinate isomorphism, I would just take the second coordinate function and show why that must be bijective and a homomorphism, that might be easier than dealing with first isomorphism theorem stuff.
I was not sure you were allowed to assume that or not, so I went that route instead.
I think it is not so easy to prove.
if (g1,h1)(G,e)=(g2,h2)(G,e), then (g1g2^(-1),h1h2^(-1))is in (G,e). Thus h1=h2,g1g2^(-1) is in G. Note that ,we cannot say g1=g2, right?
then we shall prove (f1(g1,h1),f2(g1,h1))(G,e)=(f1(g2,h2),f2(g2,h2))(G ,e)
this will require f2(g1g2^(-1),e)=e. do you mean f2(g,e)=e???why?
if we remove the condition "finitely generated", the claim in the problem will no longer be true. for example let be the direct product of infinitely many copies of and let and
the problem can be reduced to "finite" abelian groups. the reason is that we have where are some integers and are
some finite abelian groups. then gives us and see if you can solve the problem now!
anyway, to solve your problem, you should follow the idea that i gave you in my previous post. as i said, we may assume that are finite abelian. now since every finite abelian group is a
product of cyclic subgroups, we'll get another reduction: we may assume that is cyclic.
I am going to take a shot at trying to understand what you are asking.
Are you asking us whether or not the representation in terms of products of cyclic groups is unique?
In general the answer is no as we have given several examples
if and only if .
To rectify this question of multiple representations of the same group, we use invariant factor or elementary divisor decomposition, along with some division requirements.
For any finitely generated abelian group, we have
Then the free rank r and the invariant factors are uniquely determined.
Similarly, there is a uniqueness to the elementary divisor decomposition too, and it comes from essentially decomposing the elementary divisors using the isomorphism I first mentioned above into powers of primes. Then this list too is unique, up to rearranging the order of the primes.
where is a power of a prime where the are not necessarily distince primes.
You could make this unique too, by saying list the largest primes first, within those primes list the highest exponent first and then go on down until you have exhausted the list.
This is if two groups are isomorphic, then then must have the same free rank and the same invariant factor decomposition and elementary divisor decomposition (this one up to reordering).
If they do not have the same free rank, or have different invariant factor decomposition or have different elementry divisor decompositions, then they are not isomorphic.