countable group, uncountably many distinct subgroup?

Question

I need to know whether the following statement is true or false?

Every countable group $G$ has only countably many distinct subgroups.

I have not gotten any counter example to disprove the statement but an vague idea to disprove like: if it has uncountably many distinct subgroup then It must have uncountable number of element?

Have you considered a countable direct product of countable groups? — user108903, Dec 12 '12 at 16:23
The examples given in the answers prove distinct but not non-isomorphic. This is still true, and is true for certain groups with a single defining relator, by a paper of G. Baumslag and Miller (groups with a single defining relator are very natural). — user1729, Dec 14 '12 at 12:10
(On the other hand, it is not true that a countable group can have uncountably many finitely generated subgroups. You might be interested in this MO question.) — user1729, Dec 14 '12 at 12:13
@Shaun The full citation is G. Baumslag and C. F. Miller, A remark on the subgroups of finitely generated groups with one defining relation, Illinois J. Math., Volume 30, Issue 2 (1986), 255-257 MR840124 — user1729, Dec 11 '18 at 16:17

score 64 · Answer 1 · answered Dec 12 '12 at 17:13

64

Let $(\mathbb{Q},+)$ be the group of the rational numbers under addition. For any set $A$ of primes, let $G_A$ be the set of all rationals $a/b$ (in lowest terms) such that every prime factor of the denominator $b$ is in $A$. It is clear that $G_A$ is a subgroup of $\mathbb{Q}$, and that $G_A = G_{A'}$ iff $A = A'$. Since there are uncountably many sets of primes, this produces uncountably many distinct subgroups of the countable group $\mathbb{Q}$.

answered Dec 12 '12 at 17:13

Nate Eldredge

101,664

11

I like this example. In addition it shoulws that a countable ring can have uncountably many subrings. – Marc van Leeuwen Dec 14 '12 at 13:18
5

I like that this also shows that a countable group can have uncountably many distinct and non-isomorphic subgroups. Interestingly this extends to show that we can put continuum-many non-isomorphic group operations on a countable set. – Aug 17 '14 at 20:10
1

@Bryan: Hmm, is it obvious that these subgroups are non-isomorphic? – Nate Eldredge Aug 17 '14 at 20:21
4

Given two distinct collections of primes $A$ and $A'$, let's say $A$ contains $p$ while $A'$ does not. Then $G_A$ is $p$-divisible while $G_{A'}$ is not. – Aug 17 '14 at 20:26
@Bryan: Indeed, thank you. – Nate Eldredge Aug 17 '14 at 20:30
1

@Robert I'm not sure I follow. What does $p$-divisible mean here? – Anu Feb 05 '18 at 12:21
2

@Anu for every element $x$ of $G_A$ there is an element $y$ of $G_A$ such that $y^p=x$, or, in this case, $py=x$. – Feb 05 '18 at 14:04
@NateEldredge WHy sets of prime are uncountable ? Please can you explain – Curious student Nov 27 '18 at 10:08
@Shubham: There are infinitely many primes, and the power set of any infinite set is uncountable (Cantor's theorem). – Nate Eldredge Nov 27 '18 at 15:20
Thanks a lot I don't know about that – Curious student Nov 27 '18 at 15:30
1

@Shubham: See https://en.wikipedia.org/wiki/Cantor%27s_theorem. It's the same result that tells you that $\mathbb{R}$ is uncountable. – Nate Eldredge Nov 27 '18 at 15:31

score 56 · Accepted Answer · answered Dec 12 '12 at 16:23

56

One example is the group consisting of all finite subsets of $\mathbb N$, with the group operation being symmetric difference. The group is countably infinite, but for each finite or infinite $A\subseteq \mathbb N$ there's a subgroup consisting of the finite subsets of $A$.

answered Dec 12 '12 at 16:23

hmakholm left over Monica

291,611

+1. This is "the same" example as Clive Newstead's, but I think it is "more evident" to follow. – Hagen von Eitzen Dec 12 '12 at 16:26
not getting in head, could it be elaborated a little? – Myshkin Dec 12 '12 at 16:26
1

@Kuttus Which part gets stuck? That the main group is countable? Or that each subset produces a different subgroup? Or that there are uncountably many subsets of $\mathbb N$? – Hagen von Eitzen Dec 12 '12 at 16:27
well, why it is countably infinite group as you say $\mathbb{N}$ has uncountably many subsets? – Myshkin Dec 12 '12 at 16:37
Isn't that an answer to a different question? Wouldn't some free product instead be the answer? My knowledge of free products is rudimentary, but I find this question interesting. – gnometorule Dec 12 '12 at 16:51
@gnometorule: Which different question do you think this answers? You can also construct a counterexample using free groups, but that doesn't mean the one I give here fails to work. – hmakholm left over Monica Dec 12 '12 at 17:01
I think OP means to find an uncountable subgroup. Your (nice) example has a finite (not uncountably) infinite sub group. Maybe I'm just not getting it. :) – gnometorule Dec 12 '12 at 17:05
@gnometorule: To me, the question clearly speaks of the number of subgroups, not the size of any particular subgroup. – hmakholm left over Monica Dec 12 '12 at 17:08
2

You are right, and I need a refresher course in proper reading. – gnometorule Dec 12 '12 at 17:10
@Henning Makholm how each infinite subset produces different subgroup? Do you mean We have to take all singleton sets containing every element of infinite subset and all possible combinations of them? – ramanujan Nov 13 '18 at 21:22
2

@ramanujan: How would they produce identical subgroups? If $A\ne B$ then there is an $x$ that is either in $A\setminus B$ or in $B\setminus A$, so ${x}$ would only be in one of the subgroups (and therefore in particular they are not the same subgroup). – hmakholm left over Monica Nov 13 '18 at 21:24
Sir, I can see that, group is countably infinite but unable to see there are uncountably many subgroups! Further, in your answer after stating that group is countably infinite. Why you considered subset $A$ (finite or infinite) of $\mathbb{N}$? As group consist only finite subsets of $\mathbb{N}$ then why we should consider infinite subset? – Akash Patalwanshi Oct 22 '23 at 14:37
@AkashPatalwanshi Corresponding to each subset $A$ of $\mathbb N$, you get a (distinct) subgroup of the group of finite subsets of $\mathbb N$ as finite subsets of $A$ are also finite subsets of $\mathbb N$. Since there are uncountably many subsets of $\mathbb N$, you have uncountably many subgroups the main group. – Aman Kushwaha Nov 08 '24 at 18:19

Paweł Piwek · Answer 3 · 2019-07-10T14:38:58.307

There is another one: $F_\infty$, the free group on countably many letters $x_1,x_2, \ldots$.

It is countable because each of its elements is a finite string of symbols from a countable alphabet. At the same time, for every subset $A\subset \mathbb{N}$, there is a subgroup $H_A$ generated by the set $\{x_i | i \in A\}$. These subgroups are distinct, so we get uncountably many of them.

Also, $F_\infty$ embeds into $F_2$, so we can get these uncountably many subgroups within a finitely presented group!

countable group, uncountably many distinct subgroup?

3 Answers3

Linked