nonabelian group

A group is said to be nonabelian, or noncommutative, ifhas elements which do not commute, that is, if there exist $a$ and $b$in the group such that $ab\ne ba$. Equivalently, a group isnonabelian if there exist $a$ and $b$ in the group such that thecommutator $[a,b]$ is not equal to the identity of the group. Thereexist many natural nonabelian groups, with order as small as $6$.While any group for which the square map is a homomorphism is abelian,there exist nonabelian groups of order as small as $27$ for which thecube map is a homomorphism.

In the first section we give a way to visualize the group of rotationsof a sphere and prove that it is nonabelian. This should be readableby an undergraduate student in algebra. In the second section, wediscuss groups admitting a cube map and show that there are smallnonabelian examples. The second section is somewhat more technicalthan the first and will require more facility with group theory,especially working with finitely presented groups and the commutatorcalculus.

1 Concrete examples of nonabelian groups

Although most number systems we use are abelian by design, there existquite natural nonabelian groups. Perhaps the simplest example tovisualize is given by the group of rotations of a sphere. We cancompose two rotations by performing them in sequence, and we caninvert a rotation by rotating in the opposite direction, so rotationsdo form a group. To follow what rotation does to the sphere, imaginethat inside is suspended a copy of Marshall Hall’s classic text The Theory of Groups. We will keep track of three pieces ofinformation, namely, the directions that the front cover, the spine, andthe bottom of the book face. When the sphere is in the identity position,the front cover faces the reader, the spine faces the left, and thebottom of the book is oriented downward.

In preparation for verifying that the group is not abelian, we definetwo rotations, $F$ and $R$. First, let $F$ (for “flip”) be therotation which takes the point at the very top of the sphere and movesit forward through an angle of $\pi$. For example, if we start withthe sphere in the identity position and then perform $F$, the frontcover will face away from the reader, the spine will remain to theleft, and the bottom of the book will be oriented upward. Second, let$R$ (for “rotate”) be the rotation which takes the point at thevery top of the sphere and moves it left through an angle of$\frac{\pi }{2}$. If we start with the sphere in the identity positionand then perform $R$, the front cover will continue to face thereader, the spine will face downward, and the bottom of the book willbe oriented to the right.

We now verify that the group of rotations is not abelian. If we startwith the sphere in the identity position and perform $FR$, that is,first $F$, then $R$, then the front cover will face away from thereader, the spine will face downward, and the bottom will be orientedto the left. On the other hand, if we start with the sphere in theidentity position and perform $RF$, then while the front cover willface away from the reader, the spine will face upward, and thebottom will be oriented to the right. So it matters in whichorder we perform $F$ and $R$, that is, $FR\ne RF$, proving that thegroup is not abelian.

Since every rotation in three-dimensional Euclidean space can bedecomposed as a finite sequence of reflections and rotations in theEuclidean plane, one might hope that we can find finite nonabeliangroups arising from objects in the plane, and in fact we can. Foreach regular polygon, there is an associated group, the dihedral group$D_{2n}$, which is the group of symmetries of the polygon. (Here $n$denotes the number of the sides of the polygon, and $2n$ gives thenumber of elements of the group of symmetries.) It is generated bytwo elements, $F$ (for “flip”) and $R$ (for “rotate”). Theseelements can be defined by analogy with the $F$ and $R$ above; forfull details, consult the entry “Dihedralgroup (http://planetmath.org/DihedralGroup)”, where flips are labelled instead by $M$ (for“mirror”). If $n\ge 3$ (so we are dealing with an actual polygonhere), it is possible to show that $FR\ne RF$. Moreover, every groupwith order $1$, $p$, or $p^2$, where $p$ is a prime, is abelian. Thusthe smallest possible order for a nonabelian group is $6$. But$D_{2\cdot 3}$ has $6$ elements and is nonabelian, so it is thesmallest possible nonabelian group.

2 Small nonabelian groups admitting a cube map

When we say that a group admits $x\mapsto x^n$, we mean that thefunction $\phi$ defined on the group by the formula $\phi (x)=x^n$ is a homomorphism, that is, that is, that for any $x$ and $y$ inthe group,

$${(xy)}^n=\phi (xy)=\phi (x)\phi (y)=x^n{y}^n.$$

If a group admits $x\mapsto x^2$, then for any $x$ and $y$ we havethat ${(xy)}^2=x^2{y}^2$. Multiplying on the left by $x^{-1}$ and onthe right by $y^{-1}$ yields the identity $yx=xy$. Thus all suchgroups are abelian. Moreover, the generalized commutativity andassociativity laws for abelian groups imply that an abelian groupadmits all maps $x\mapsto x^n$. It is therefore reasonable to wonderwhether the converse holds. In fact it is possible for a nonabeliangroup to admit $x\mapsto x^3$. The smallest order for such a group is$27$. It is beyond the scope of this entry to prove that $27$ is the smallestpossible order, but we will give an explicit example.

Let $G$ be the group with presentation

$$G=⟨a,b,c\mid a^3,b^3,c^3,[a,c],[b,c],[a,b]c⟩.$$

This can be realized concretely as the group of upper-triangularmatrices over $\mathbb{Z}/3\mathbb{Z}$ with $1$s on the diagonal, but forsimplicity we shall work directly with the presentation.

The first three relators tell us that each generator of the group hasorder $3$. The next two tell us that $c$ is central — since itcommutes with the other two generators and commutes with itself, itmust therefore commute with everything. The final relator is perhapsthe most interesting. We can interpret it as the rewrite rule

$$ba\mapsto abc,$$

that is,

“when $b$ moves past $a$ it turns into $bc$.”

Thus given an element of $G$ we can always write it in the normal form$a^j{b}^k{c}^{\mathrm{\ell }}$, where , and all such elementsare distinct. This proves that the cardinality of $G$ is $27$.Moreover, we also observe that

$$ba=abc\ne ab,$$

so $G$ is not abelian.

It remains to check that $G$ admits the cube map. We will prove thesimpler statement that $G$ has exponent $3$. Given $x$ in $G$, wefirst normalize it, so $x=a^j{b}^k{c}^{\mathrm{\ell }}$. Since $c$ is in thecenter of $G$,

$$x^3={(a^j{b}^k)}^3{c}^{3\mathrm{\ell }}={(a^j{b}^k)}^3=a^j{(b^k{a}^j)}^2{b}^k.$$

To normalize the word $b^k{a}^j$, we push each $b$ past all of the$a$s. Since pushing $b$ past a single $a$ turns it into $bc$, pushingit past $a^j$ turns it into $b{c}^j$, that is,

$$b^k{a}^j=b^{k-1}{a}^{j}b{c}^j.$$

By induction it follows that

$$b^k{a}^j=a^j{b}^k{c}^{jk}.$$

Applying this result to $x^3$, we get that

$$x^3=a^j{(a^j{b}^k{c}^{jk})}^2{b}^k=a^{2j}(b^k{a}^j)b^{2k}{c}^{2jk}=a^{3j}{b}^{3k}{c}^{3jk}=1.$$

Since $x^3$ is trivial for any $x$, it follows that $G$ admits thecube map.

The other nonabelian group of order $27$ has exponent $9$ and also admits thecube map. This will be described in an attached entry.