If $G$ is an infinite group then $G$ has infinitely many subgroups.

Proof: Let's consider the following set: $C=\{\left \langle g \right \rangle: g\in G \}$ - collection of all cyclic subgroups in $G$ generated by elements of $G$. Two cases are possible:

  1. Exists infinitely many distinct cyclic subgroups $\Rightarrow$ We are done.

  2. Exists finitely many distinct cyclic subgroups for example $C=\{H_1, H_2,\dots, H_n\}$. Then $G=\bigcup \limits_{i=1}^{n}H_i$. Since $G$ is infinite then WLOG suppose that $H_1$ is also infinite, where $H_1=\left \langle g_1 \right \rangle$. Let's consider the following set $\{\left \langle g_1^n \right \rangle: n\in \mathbb{N}\}$ - the collection of all cyclic sugroups of $H_1\subset G.$ Let $K_1=\left \langle g_1 \right \rangle$, $K_2=\left \langle g_1^2 \right \rangle$, $K_3=\left \langle g_1^3 \right \rangle$, $\dots$. It's easy to show that $K_n$ and $K_m$ are distinct for $n\neq m$. Indeed, WLOG take $n<m$ and taking $g_1^n\in K_n$ but $g_1^n\notin K_m$ otherwise $g_1^n=g_1^{ml}$ where $l\in \mathbb{Z}$ $\Rightarrow$ $g_1^{n-ml}=e$ and since $H_1$ is infinite $\Rightarrow$ $n=ml$ which is contradiciton since $m>n$.

Thus, the subgroups $K_n$ for any $n\in \mathbb{N}$ are cyclic subgroups of $H_1$ $\Rightarrow$ cyclic subgroups of $G$.

Is this reasoning correct?

  • 1
    $\begingroup$ I would add a treatment of the case where one of the cyclic subgroups is itself infinite like $\mathbb Z$. $\endgroup$ – hardmath Nov 30 '17 at 16:30
  • $\begingroup$ @hardmath, the case when one of the subgroups is infinite i have already considered in 2 ($H_1$ is infinite). $\endgroup$ – ZFR Nov 30 '17 at 17:01
  • $\begingroup$ @hardmath, honestly speaking i do not understand your comment. Could you clarify your remark? $\endgroup$ – ZFR Nov 30 '17 at 17:09
  • $\begingroup$ I'll post an Answer showing how I'd pull out the "special case" for clarity. $\endgroup$ – hardmath Nov 30 '17 at 17:14
  • $\begingroup$ Every group is the union of the generators of its elements. $\endgroup$ – monoidaltransform May 8 '19 at 21:22

I think the contrapositive is much clearer:

If a group has finitely many subgroups, then the group is finite.

Indeed, let $G$ be a group with finitely many subgroups. Then $G$ has finitely many cyclic subgroups. An infinite cyclic group has infinitely many subgroups. Therefore, all cyclic subgroups of $G$ are finite. Finally, $G$ is finite because it is the union of its cyclic subgroups, which is a finite union of finite sets.


The proof given is correct, and I'm suggesting an alternative only for the sake of style/clarity (which is more subjective than correctness).

The point in the OP's proof where a detailed argument appears is nested inside the case analysis (finitely many vs. infinitely many cyclic subgroups). Pulling that argument out as a Lemma serves both to motivate the result and to simplify the main argument that follows:

Lemma An infinite cyclic group has infinitely many (cyclic) subgroups.

Proof: An infinite cyclic group is isomorphic to additive group $\mathbb Z$. Each prime $p\in \mathbb Z$ generates a cyclic subgroup $p\mathbb Z$, and distinct primes give distinct subgroups. So the infinitude of primes implies $\mathbb Z$ has infinitely many (distinct) cyclic subgroups. QED

Proposition An infinite group has infinitely many (cyclic) subgroups.

Proof: Let $G$ be an infinite group. Every $g\in G$ belongs to at least one cyclic subgroup of $G$, namely $\langle g \rangle$. (1) If there exist infinitely many (distinct) cyclic subgroups of $G$, then we are done.

So assume (2) $G$ has only finitely many cyclic subgroups $H_1,H_2,\ldots,H_k$. Since $G$ is infinite, at least one of these $H_i$ must be infinite (otherwise we have a finite covering of $G$ with finite sets, implying $G$ is finite). Then the Lemma above says such infinite $H_i$ has infinitely many cyclic subgroups, which implies also that $G$ does (since a cyclic subgroup of $H_i$ is a cyclic subgroup of $G$). QED

Assumption (2) actually leads to a contradiction, but we haven't highlighted that. Some authors would prefer to phrase the proof in those terms, but I wanted to emphasize keeping your structure of proof after pulling out the case where $G$ is infinite cyclic as a Lemma.

  • $\begingroup$ Thanks for remark! +1 :) $\endgroup$ – ZFR Nov 30 '17 at 18:15
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    $\begingroup$ No need to use that there are infinitely many primes: the subgroups $n\mathbb Z$ for each $n \in \mathbb N$ are all distinct because they have distinct indices. $\endgroup$ – lhf Dec 1 '17 at 1:04
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    $\begingroup$ @lhf: I never miss an opportunity to appeal to the "infinitude of primes". $\endgroup$ – hardmath Dec 1 '17 at 2:04
  • $\begingroup$ @hardmath: Sorry if this is too simple, but why must a group be contained in the union of its subgroups? $\endgroup$ – MSIS Nov 13 '20 at 17:39
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    $\begingroup$ @MSIS: I'm actually using a slight stronger claim, that a group is contained in the union of its cyclic subgroups (the subgroups generated by some particular group element). That is, if $x\in G$ is a particular group element, $x \in \langle x \rangle$, the cyclic subgroup of $G$ generated by $x$. If $G$ itself is not cyclic, then $\langle x \rangle$ must be a proper subgroup. But if $G$ is cyclic, it's possible that $x$ would generate all of $G$. $\endgroup$ – hardmath Nov 13 '20 at 18:36

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