we all know that set of rationals i a subgroup of set of reals. my question is whether there exsts a group between these tw groups. f yes what it can be? and if no, how to prve the non-existence?
n_bourbaki
Jul24-08, 06:44 PM
There exist uncountably many. (We assume that you mean addition as the group law.)
morphism
Jul24-08, 10:31 PM
It will help your intuition if you notice that the reals form a 'very large' vector space over the rationals.
anugrah
Jul24-08, 11:44 PM
sorry, forgot to mention that the group law is multiplication. I wanted everything with respect to multiplication. And if there exists a subgroup between these two groups (even with respect to addition ) please give an example of the same.
d_leet
Jul25-08, 12:55 AM
sorry, forgot to mention that the group law is multiplication. I wanted everything with respect to multiplication. And if there exists a subgroup between these two groups (even with respect to addition ) please give an example of the same.
Pick any irrational number k and consider the set Q'={qkn | q is a nonzero rational number, and n is an integer}
It is easy to show that this is a group under multiplication, and easy to show that the nonzero rationals are a subgroup of this group, but clearly since this group is countable it is proper subgroup of the reals.
n_bourbaki
Jul25-08, 04:17 AM
sorry, forgot to mention that the group law is multiplication.
The real numbers under multiplication are not a group.
I wanted everything with respect to multiplication. And if there exists a subgroup between these two groups (even with respect to addition ) please give an example of the same.
It would be better if you tried to work it out yourself.