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I was wondering this today, and my algebra professor didn't know the answer.

Are subgroups of finitely generated groups also finitely generated?

I suppose it is necessarily true for finitely generated abelian groups, but is it true in general?

And if not, is there a simple example of a finitely generated group with a non-finitely generated subgroup?

NOTE: This question has been merged with another question, asked by an undergraduate. For an example not involving free groups, please see Andreas Caranti's answer, which was the accepted answer on the merged question.

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    $\begingroup$ Like you noted, this is indeed true for finitely generated abelian groups. Put another way, this is the same as the statement that every finitely generated Z-module is Noetherian, which holds since Z is Noetherian. $\endgroup$
    – Jason Juett
    Jul 9, 2016 at 11:33
  • $\begingroup$ Since there are Noetherian groups, you can guess the answer is no. $\endgroup$
    – Qian
    Jul 31, 2016 at 15:54

5 Answers 5

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No. The example given on Wikipedia is that the free group $F_2$ contains a subgroup generated by $y^n x y^{-n}, n \ge 1$, which is free on countably many generators.

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    $\begingroup$ But wouldn't that subgroup still be generated by $x,y$? $\endgroup$
    – GuPe
    Jun 29, 2017 at 15:56
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    $\begingroup$ @Guacho: no. $y$ is not an element of the subgroup. $\endgroup$
    – Qiaochu Yuan
    Jun 29, 2017 at 23:32
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It is well-known that the free group $F_2$ on two generators has as a subgroup a group isomorphic to a free group on a countably infinite set of generators. See Qiaochu's example.

However a finite index subgroup of a finitely generated group is finitely generated.

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A theorem of Higman, Neumann, and Neumann says that every countable group (no matter what horrible properties it might have) can be embedded as a subgroup of a group generated by $2$ elements. Thus subgroups of finitely generated groups can be pretty much anything.

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One of the easiest (counter)example is in Hungerford's Algebra.

Let $G$ be the multiplicative group generated by the real matrices $$a = \left(\begin{array}{l l} 1 & 1\\ 0 & 1 \end{array}\right), b = \left(\begin{array}{l l} 2 & 0\\ 0 & 1 \end{array}\right) $$ Let $H$ be the subgroup of $G$ consisting of matrices that have $1$s on the main diagonal. Then $H$ is not finitely generated.

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    $\begingroup$ isn't H just generated by a? $\endgroup$
    – fdzsfhaS
    Jul 12, 2016 at 19:20
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    $\begingroup$ @DavidWarrenKatz The powers of $a$ give only integers on the $(1,2)$ entry, while $H$ has on that entry rational numbers of the form $p/2^q$. $\endgroup$
    – user26857
    Jun 2, 2019 at 21:33
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    $\begingroup$ @mez Why $H$ is not finitely generated? $\endgroup$
    – M.Ramana
    May 10, 2020 at 5:31
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Perhaps an elementary example can be provided by the wreath product of two copies of (the additive group of) $\mathbf{Z}$.

Take copies $G_{i}$ of $\mathbf{Z}$, for $i \in \mathbf{Z}$, and let $$ B = \coprod_{i \in \mathbf{Z}} G_{i} $$ be the direct sum (coproduct in the category of abelian groups).

Now let another copy $H = \langle h \rangle$ of $\mathbf{Z}$ act on $B$ by $$ G_{i}^{h} = G_{i+1}. $$ More precisely, conjugation by $h$ takes a generator $g_{i}$ in the copy $G_{i}$ of $\mathbf{Z}$ to a generator $g_{i+1}$ of the $(i+1)$-th copy.

Then the semidirect product $G = B \rtimes H$ is generated by $g_{0}$ and $h$, but its subgroup $B$ requires an infinite number of generators.

It is easy to see what is going on. $B$ requires an infinite number of generators $g_{i}$. Now $h$ takes one of these generators by conjugation to all others.

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    $\begingroup$ OK, turns out that this one works! I wouldn't have thought someone would know wreath products before free groups, but I guess it's understandable they might. $\endgroup$
    – Tara B
    Feb 19, 2013 at 17:15
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    $\begingroup$ @TaraB, thanks! I was exposed to this very construction (in the elementary terms I have tried to use above, and without any mention of the term wreath product) before I learned about free groups. The key concept is that of a semidirect product. $\endgroup$
    – Andreas Caranti
    Feb 19, 2013 at 17:21
  • $\begingroup$ Yes. I don't think I even encountered semidirect products until after free groups, myself, though. In fact I know I didn't, because I definitely knew about presentations by then. $\endgroup$
    – Tara B
    Feb 19, 2013 at 17:23
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    $\begingroup$ Note that this answer and these comments were made in a different context (this question was merged with another question asked by an undergraduate who hadn't encountered free groups). $\endgroup$
    – Tara B
    Feb 19, 2013 at 18:14
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    $\begingroup$ @exfret, thanks, I have added a line to clarify that I meant the coproduct in the category of abelian groups. $\endgroup$
    – Andreas Caranti
    Feb 15, 2019 at 13:11

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