I wanted to know how many distinct number of subgroups are there for a group containing countable number of elements. Z under addition is countable and it has countable number of subgroups.
Let me give you another example. Take the Abelian group of the rationals \Q. For any family P of primes, there is a subgroup Z[1/P]:=Z[1/p :p\in P] of \Q, which is the group of all rationals whose denominator can be written as product of elements in P (this is really the subring of \Q generated by 1 and all the 1/p).
Of course, Z[1/P] \neq Z[1/P'] whenever P\neq P'.
To conclude just notice that the power set of the set of primes is not countable.
Excuse me, Simone, but the first example given by Hans was also about an Abelian group.
Indeed, let Fin(N) be a family of all finite subsets of the set N of natural numbers.
Denote by ^ the XOR (symmetric difference) binary operation,
where A^B=(A\cup B)/(A\cap B) for any A, B\in FIN(N).
Obviously, ^ is an accociative and commutative operation, and \emptyset is a neutral element, moreover, A^A=\emptyset for any A.
Therefore, (FIN(N), ^) is an Abelian group.
Further, the set FIN(N) is countable itself, and for each its subset S the subset S \cup {\emptyset} be distinct subgroup. Therefore the group FIN(N) has uncountable set of subgroups due to well known continuum hypothesis.
Uncountability of the power set of N has nothing to do with the continuum hypothesis. It is just Cantors old result that the power set of a set X has larger cardinality than the set X itself.
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