Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|cite|improve this question
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.… – Alex Youcis Mar 12 '13 at 16:41

6 Answers 6

up vote 16 down vote accepted

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.

share|cite|improve this answer

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 $3$).

share|cite|improve this answer
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 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.

share|cite|improve this answer

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 $G$.

share|cite|improve this answer

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

share|cite|improve this answer

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.

share|cite|improve this answer
whoops, this is a copy of Brian's answer. – Ben Sprott Mar 12 '13 at 18:07

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.