# Number of elements in a group and its subgroups (GS 2013)

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?

• Well, $(\Bbb R,+)$ won't get you anywhere, since it is uncountable. – Cameron Buie Nov 7 '13 at 15:43
• It certainly does. Each non-negative $\alpha\in\Bbb R$ generates a distinct cyclic subgroup $\langle\alpha\rangle,$ for example, and there are uncountably-many such $\alpha$. – Cameron Buie Nov 7 '13 at 16:10
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.
• Thank you for your answer but I am not strong enough in algebra. Please expand the answer such that I can grasp it. "sum of copies of the group $\frac{\mathbb{Z}}{2\mathbb{Z}}" - not clear. – Dutta Nov 7 '13 at 16:13 • Are you familiar with the notion of the sum of finitely many groups? – Mikhail Katz Nov 9 '13 at 20:03 • It is clear now. I have studied recently direct product of groups a little. – Dutta Nov 10 '13 at 2:58 I think$\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$. This does the trick. • such$S$has to be finite no? can it be infinite? – GA316 Nov 7 '13 at 16:53 • With finite$S$'s you won't get an uncountable collection of subgroups, so you have to allow$S$to be infinite. – Mikhail Katz Nov 7 '13 at 17:26 This wont answer to your question, but still it is interesting. consider$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}}$has uncountably many subgroups. • Thank you for your answer. This is one more example of uncountable group whose collection of subgroups is uncountable. – Dutta Nov 8 '13 at 3:13 • Where are you studying? – Dutta Nov 8 '13 at 3:14 • @Samprity IMSc, Chennai – GA316 Nov 8 '13 at 4:37 • Grate! So you are very meritorious and intelligent. – Dutta Nov 8 '13 at 10:20 Take$\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.