# Examples of abelian subgroups of non-abelian groups.

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

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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/… –  Alex Youcis Mar 12 '13 at 16:41

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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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. –  user1729 Mar 12 '13 at 17:05

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 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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