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I'm searching for examples of abelian subgroups of non-abelian groups. Please enlighten me.

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    $\begingroup$ It may interest you to know that there is, at an extreme side of things, a characterization of (finite)non-abelian groups for which every proper subgroup is abelian. groupprops.subwiki.org/wiki/… $\endgroup$ – Alex Youcis Mar 12 '13 at 16:41
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Let $G$ be any group, Abelian or not, and let $g\in G$. Then $\langle g\rangle=\{g^n:n\in\Bbb Z\}$, the subgroup of $G$ generated by $g$, is Abelian.

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The smallest non-abelian group is the symmetric group $S_3$ of order $6$. So all its proper subgroups are abelian (the trivial subgroup, three subgroups of order $2$ and one subgroup of order $3$).

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  • $\begingroup$ This is also true for any groups of order $pq$ where $p$ and $q$ are primes, as any proper, non-trivial subgroup will have order $p$ or order $q$ and so be cyclic. $\endgroup$ – user1729 Mar 12 '13 at 17:05
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The more simple example is the trivial subgroup $\{e\}$ of any non-abelian group where $e$ is the identity element.

Another example less simple is the quaternion group and for example $\{\pm1\}$ is an abelian subgroup.

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The dihedral groups are examples of non-abelian groups with a maximal subgroup which is cyclic, thus abelian.

Other examples of the same kind can be constructed by taking two primes $p, q$, and consider the order $n$ of $q$ modulo $p$, so the smallest $n$ such that $q^{n} \equiv 1 \pmod{p}$.

Then the finite field $\mathbf{F}_{q^{n}}$ contains an element $g$ of multiplicative order $p$. If we consider the semidirect product $G$ of $A$, the additive group of $\mathbf{F}_{q^{n}}$, by $\langle g \rangle$, acting by multiplication, then $A$ will be an abelian, maximal subgroup of the non-abelian group $G$.

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The center of the group is a subgroup. It doesn't necessarily contain all Abelian subgroups.

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One nice example is to look at SU(2). Given two elements randomly from SU(2), they will likely not commute and will thus generate a non-abelian group under free product. However, the subgroups formed under free product of each of the two elements alone do form abelian groups and these groups are naturally subgroups of the one generated by the two elements.

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  • $\begingroup$ whoops, this is a copy of Brian's answer. $\endgroup$ – Ben Sprott Mar 12 '13 at 18:07
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Every non Abelian group has a nontrivial Abelian subgroup. And Every nontrivial abelian group has a cyclic subgroup.

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Let G be a group having order p^3 where p is a prime. Then its proper subgroup can have order either 1 or p or p^2. We know that a group of prime order is cyclic and hence it is abelian. Also we know that a group of order p^2 is abelian. Thus each of the proper subgroup of G is abelian. Even though G may be non abelin. For example The group of Quatarians is of order 8, and each of its proper subgroup is abelian even though group itself is not abelian.

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