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?

  • $\begingroup$ Have you considered a countable direct product of countable groups? $\endgroup$ – user108903 Dec 12 '12 at 16:23
  • 1
    $\begingroup$ 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). $\endgroup$ – user1729 Dec 14 '12 at 12:10
  • 2
    $\begingroup$ (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.) $\endgroup$ – user1729 Dec 14 '12 at 12:13
  • 1
    $\begingroup$ @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 $\endgroup$ – user1729 Dec 11 '18 at 16:17

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 $A$.

  • $\begingroup$ +1. This is "the same" example as Clive Newstead's, but I think it is "more evident" to follow. $\endgroup$ – Hagen von Eitzen Dec 12 '12 at 16:26
  • $\begingroup$ not getting in head, could it be elaborated a little? $\endgroup$ – Marso Dec 12 '12 at 16:26
  • 1
    $\begingroup$ @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 $\mathbb N$? $\endgroup$ – Hagen von Eitzen Dec 12 '12 at 16:27
  • 1
    $\begingroup$ You are right, and I need a refresher course in proper reading. $\endgroup$ – gnometorule Dec 12 '12 at 17:10
  • 1
    $\begingroup$ @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). $\endgroup$ – Henning Makholm Nov 13 '18 at 21:24

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 $\mathbb{Q}$.

  • 5
    $\begingroup$ I like this example. In addition it shoulws that a countable ring can have uncountably many subrings. $\endgroup$ – Marc van Leeuwen Dec 14 '12 at 13:18
  • 4
    $\begingroup$ 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. $\endgroup$ – Robert Wolfe Aug 17 '14 at 20:10
  • 2
    $\begingroup$ 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. $\endgroup$ – Robert Wolfe Aug 17 '14 at 20:26
  • 1
    $\begingroup$ @Robert I'm not sure I follow. What does $p$-divisible mean here? $\endgroup$ – Anu Feb 5 '18 at 12:21
  • 2
    $\begingroup$ @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, $py=x$. $\endgroup$ – Robert Wolfe Feb 5 '18 at 14:04

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!


A very natural way to go from countably infinite set to uncountable set is to take its power set (set of all subsets).

Let $X=\{p_1,p_2,\cdots\}$ be the set of all primes. Since we can count the primes as “first prime is $2$, second is $3$, third is $5$,…” this set is a countably infinite set. (Other way of seeing this is $X$ is an infinite subset of $\mathbb N$.)

The power set of $X$, $\mathcal P (X)$ is therefore an uncountable set.

The natural way subgroups come to the mind is cyclic groups generated by $1/\pi$. But if we only take this collection we will end up with as many groups as there are primes, which will not be helpful.

But let us take subgroup generated by a collection of $1/\pi$’s. That will take us to the set of subsets of $X$, which will give what we desire.

Let us make this more formal:

Consider a subset $A$ of $X$. $GA$ be the subgroup generated by $\{1/p\alpha \mid p\alpha \in A\}$ The only obstacle that can occur is that these $GA$s are not distinct for distinct $A$s. In other words, the function from $\mathcal P (X)$ to subgroups of $\mathbb Q$ which takes a set $A$ to $GA$ must be one-to-one.

Let $GA=GB$ We wish to show that $A\subseteq B$ and $B\subseteq A$.

Let $p\in A$. Then $1/p\in GA$. Then $1/p\in GB$. Now if $p\not \in B$ then there is no way that $1/p$ comes in $GB$, because any linear combination of $1/p\alpha$s where $p\alpha$ is different from $p$, will never give $p$ in the denominator (because of the lcm process in addition of two rationals). Therefore, $p\in B$. This proves that $A\subseteq B$. Since there is nothing special about $A$ and $B$, we can interchange the roles of these two, and finally conclude that $A=B$.

Therefore for each subset of $X$, we get a different subgroup of $\mathbb Q$, and since there are uncountable number of subsets of $X$, we get an uncountable number of subgroups of $\mathbb Q$.

New contributor
Harry Richie is a new contributor to this site. Take care in asking for clarification, commenting, and answering. Check out our Code of Conduct.
  • $\begingroup$ The answer is unintelligible until we suddenly are informed that you are working in $(\mathbb{Q},+)$ at the very end of the answer. $\endgroup$ – Arturo Magidin yesterday

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.