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I'm looking for a non-abelian group which has infinitely many abelian subgroups. Do you know any examples of such groups?

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    $\begingroup$ Doesn't a direct product construction get you an example easily? $\endgroup$
    – hardmath
    Commented Jun 17, 2016 at 1:32
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    $\begingroup$ Perhaps the "minimal" example in some sense is the infinite dihedral group. $\endgroup$
    – hardmath
    Commented Jun 17, 2016 at 1:43

4 Answers 4

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Any infinite group $G$ must have infinitely many abelian subgroups. Note that for each $x \in G$, there is a cyclic subgroup $\langle x \rangle$, which is abelian. If there is an $x$ such that $\langle x \rangle$ is infinite, then $\langle x \rangle$ has infinitely many abelian subgroups. If no such $x$ exists, there must be infinitely many distinct finite cyclic subgroups $\langle x \rangle$, since otherwise $G$ would be the finite union of finite sets.

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    $\begingroup$ Remarkably, it is impossible not to find an example for the OP question. - Rewording your argument: Each element of $G$ generates an abelian subgroup, and for each such subgroup there are only finitely many generators, hence for infinite $G$, there must be at least $|G|$ many abelian subgroups :) $\endgroup$ Commented Jun 16, 2016 at 21:42
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    $\begingroup$ Your proof is incomplete. You must proof that there exists at least one infinite group. $\endgroup$ Commented Jun 16, 2016 at 21:44
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    $\begingroup$ @PyRulez I'd argue otherwise since the existence of such groups is fairly well-known. $\endgroup$
    – user217285
    Commented Jun 16, 2016 at 22:12
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    $\begingroup$ It doesn't prove that a cyclic (sub)group is abelian either. Or that there exists a non-abelian group, let alone an infinite non-abelian group. What do you expect in four lines of text, of course it's incomplete, but that doesn't mean it must be made complete ;-) $\endgroup$ Commented Jun 17, 2016 at 1:26
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    $\begingroup$ This question therefore immediately devolves to "list a bunch of infinite nonabelian groups." Great. $\endgroup$
    – djechlin
    Commented Jun 17, 2016 at 16:29
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Take the product $G = S_3 \times \Bbb Z$, which is non abelian since it has a non-abelian subgroup, namely $S_3$.

However, $\{1\} \times n\Bbb Z$ are abelian subgroups of $G$ for every $n \geq 0$.

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    $\begingroup$ could you please explain to me what's the operation on such group? $\endgroup$
    – Angie
    Commented Jun 16, 2016 at 17:07
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    $\begingroup$ @Angie : the direct product of $G$ and $H$ has the following operation : $(g,h) \cdot (g',h') = (gg', hh')$. $\endgroup$
    – Watson
    Commented Jun 16, 2016 at 17:09
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    $\begingroup$ So in our case, for instance $(\text{id}, 3)$ * $(\sigma, 4) = (\sigma, 7)$ where $\sigma \in S_3$ is any permutation. $\endgroup$
    – Watson
    Commented Jun 16, 2016 at 17:10
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Consider the subgroups of $\mathrm{SO}(3)$ (visualized as the rotational symmetries of the $2$-sphere) representing rotations about a fixed axis through the center of this sphere. There are infinitely many choices of this axis, each of which specifies an (abelian) subgroup isomorphic to $\mathrm{U}(1)$.

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The set of $2\times 2$ matrices with real entries is non-Abelian when the operator is multiplication, but it has an infinite number of Abelian subgroups.

For example consider any subgroup of the form $$\{A | A = \begin{bmatrix} p^n & 0\\ 0 & 1\end{bmatrix} \mbox{ where } n\in \mathbb{Z}\}$$ where $p$ is a constant and can be any prime.

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