Thread: tensor product

If G is finite Abelian group and G (X)_Z Z/pZ={0} ( tensor of G with Z/pZ over Z )
for all primes p, then show that G = {0}. Does the result remain true if G is infinite ?

Since G is finite abelian then G= Z/n_1Z x Z/n_2Z x ... x Z/n_kZ
where n_k | n_k-1 | ... | n_2 | n_1
so {0}= G(X)Z/pZ = Z/dZ where d=gcd(n_k , p) for all primes
then d=1 and for each p there is one of n_i relatively prime to p
but can this imply G is zero ?
Also, I have no idea about the second assertion.

Originally Posted by hgd7833

If G is finite Abelian group and G (X)_Z Z/pZ={0} ( tensor of G with Z/pZ over Z )
for all primes p, then show that G = {0}. Does the result remain true if G is infinite ?

Since G is finite abelian then G= Z/n_1Z x Z/n_2Z x ... x Z/n_kZ
where n_k | n_k-1 | ... | n_2 | n_1
so {0}= G(X)Z/pZ = Z/dZ where d=gcd(n_k , p) for all primes
then d=1 and for each p there is one of n_i relatively prime to p
but can this imply G is zero ?
Also, I have no idea about the second assertion.

The second assertion is easy to show is false: take , with any finite abelian group, say of order

...

Tonio

No. The question is : what if G is infinite , you're assuming G is finite in your argument.

Originally Posted by hgd7833

No. The question is : what if G is infinite , you're assuming G is finite in your argument.

No, is infinite and Z-tensored with any finite abelian group, and in particular with we get the zero group, thus showing that an infinite group can give zero tensored with any group of order a prime p and nevertheless it is not, obviously, the zero group.

Tonio

Oh yes, so Q is your group. This works now.
Did you look at my argument for the first part ? Is it correct ? Thanks

for the first part, suppose to the contrary that then and thus for all and all primes

in particular for all primes but that is false because if then

Yes I got it now. Thanks very much to you and to Tonio !