For a group G let F(G) denote the collection of all subgroups of G.Which of the following situation can occur?

  1. G is finite but F(G) is infinite.
  2. G is infinite but F(G) is finite.
  3. G is countable but F(G) is uncountable.
  4. G is uncountable but F(G) is countable.

Attempt. If G is finite then P(G) is finite hence (1) is not possible. Let G=I(the set of integers) be a group under addition then Consider S=$\{$Group of integers modulo n for n$\epsilon N$}$\subset F(G)$ hence infinite therefore (2) is incorrect.(4) is also incorrect Consider R(the set of real numbers) under addition then Let S=$\{ma|a\epsilon R$-{set of irrational numbers $m\epsilon I$}$\}$ $\subset F(G)$ where S is uncountable therefore F(G) is uncountable hence (3) is the only left choice. Why (3) is possible?

  • 2
    $\begingroup$ To do this, you can't just think of a single situation where it does or doesn't occur. You have to prove that either it can occur (a specific example works for this), or that in EVERY situation it can't occur $\endgroup$ – Mark Aug 31 '16 at 6:10
  • $\begingroup$ As for why (3) is possible, here is an example of a countable group with uncountably many subgroups. $\endgroup$ – Mark Aug 31 '16 at 6:12
  • $\begingroup$ The question doesn't say that exactly one is possible and the others aren't. Multiple could be possible or none. $\endgroup$ – fleablood Aug 31 '16 at 7:18
  • You are right that (1) is impossible.

  • For (2), your reasoning is incorrect, for two reasons: first, the integers modulo $n$ are not a subgroup of the integers, but rather a quotient group. Second, in order to prove that (2) is impossible you'd have to show that it's false for ALL groups, not just for the integers.

    You are right that (2) is impossible. To prove it, let $A$ be an infinite group and consider the subgroup generated by each element $a \in A$. There are two cases: first, one of these subgroups is infinite and isomorphic to $\mathbb{Z}$; second, all of these subgroups are finite. In the first case, you should be able to show that $\mathbb{Z}$ has infinitely many subgroups. In the second, $F(G)$ contains a finite subgroup containing each element, and finitely many finite subgroups could not possibly cover every element.

  • For (3), Mark has already linked to why it is possible in this question.

  • For (4), again you cannot prove it is impossible just by giving a single example, you have to show it is false for ALL groups. Try to employ a similar technique to how we proved (2): consider $A$ to be an uncountable group, and for each $a \in A$ there is a subgroup generated by $a$ which is at most countable. If there were countably many subgroups total, then the countable union of countable sets is countable, so...

So the only one that is possible is (3).


The arguments in (2) and (4) generalize as follows: if $G$ is infinite, then $|F(G)| \ge |G|$. But this is not true if $G$ is finite, e.g. if $G$ is cyclic of prime order then it only has two subgroups.

  • 1
    $\begingroup$ As a side remark (which I find interesting), which is beyond the scope of the question in some sense, to show that a countable union of countable sets is countable one needs some sort of axiom of choice, so 4 could be true in some settings. Of course most people work with axiom of choice, so you are completely right. $\endgroup$ – Paul Plummer Sep 1 '16 at 1:19

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