# Non-abelian group with infinitely many abelian subgroups

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

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

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.

• 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 :) Commented Jun 16, 2016 at 21:42
• Your proof is incomplete. You must proof that there exists at least one infinite group. Commented Jun 16, 2016 at 21:44
• @PyRulez I'd argue otherwise since the existence of such groups is fairly well-known. Commented Jun 16, 2016 at 22:12
• 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 ;-) Commented Jun 17, 2016 at 1:26
• This question therefore immediately devolves to "list a bunch of infinite nonabelian groups." Great. Commented Jun 17, 2016 at 16:29

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 $n \geq 0$.

• could you please explain to me what's the operation on such group? Commented Jun 16, 2016 at 17:07
• @Angie : the direct product of $G$ and $H$ has the following operation : $(g,h) \cdot (g',h') = (gg', hh')$. Commented Jun 16, 2016 at 17:09
• So in our case, for instance $(\text{id}, 3)$ * $(\sigma, 4) = (\sigma, 7)$ where $\sigma \in S_3$ is any permutation. Commented Jun 16, 2016 at 17:10

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 $\mathrm{U}(1)$.

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.