Every countable group has only countably many distinct subgroups.
The above statement is false. How to show it? One counterexample may be sufficient, but I am blind to find it out. I have considered some counterexample only like $(\mathbb{Z}, +)$, $(\mathbb{Q}, +)$ and $(\mathbb{R}, +)$.
Is there any relationship between number of elements in a group and number of its subgroup? I do not know. Please discuss a little. What will be if the group be uncountable?
Thank you for your help.
On
I think $\mathbb Q$ is also a counterexample. There are uncountably many subcollections $S$ of the collection of prime numbers. For every such $S$, consider the subgroup $G_S$ of $\mathbb Q$ consisting of numbers $m/n$ where the denominator $n$ is divisible only by prime numbers from $S$. This does the trick.
On
Take $\bigoplus_{i \in \Bbb N}\Bbb Z_i $ direct sum of countable many copies of $(\Bbb Z,+)$. It's free group with countable base. $Card(\bigoplus_{i \in \Bbb N}\Bbb Z_i)=Card(\Bbb N^*)$. It's easy to see that set of subgroups of $\bigoplus_{i \in \Bbb N}\Bbb Z_i $, $\{\bigoplus_{i \in A}\Bbb Z_i \mid A\in \mathcal P(\Bbb N)\}$ is uncountable.
$\bigoplus_{i \in \Bbb N}\Bbb Z_i $ is a subgroup of $\prod_{i \in \Bbb N}\Bbb Z_i$ consisting of those sequences which are $0$ everywhere but finite number of places. Thats why its countable. Now for different $A,B\subset \Bbb N$ we get different subgroups $\bigoplus_{i \in A}\Bbb Z_i $, $\bigoplus_{i \in B}\Bbb Z_i $. There is continuum different subset of $\Bbb N$ and we are done.
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This wont answer to your question, but still it is interesting. consider $S_{\mathbb{N}}$ , the set of all bijections from $\mathbb{N}$ to $\mathbb{N}$ which is uncountable in cardinality. Now for any non empty subset $A$ of $\mathbb{N}$ we can find a subgroup of $S_{\mathbb{N}}$ which is isomorphic to $S_A$ namely the collection of all bijections in $S_{\mathbb{N}}$ which fixes all the elements of $\mathbb{N} - A$. Hence $S_{\mathbb{N}}$ has uncountably many subgroups.
The countably infinite sum $S$ of copies of the group $G=\mathbb Z/2\mathbb Z$ (cyclic group of order two) indexed by $I$ is countable. Every subset $A$ of the index set $I$ corresponds to a subgroup of $S$ consisting of elements with nonzero components only in the copies of $G$ corresponding to the subset $A\subset I$. This gives uncountably many subgroups because the set of subsets of $I$ is uncountable.