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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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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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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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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