# Constructions of the smallest nonabelian group of odd order

I write $$|X|$$ for the number of elements in a finite set $$X$$. Recall some basic facts:

If $$p$$ is a prime number, then any group $$G$$ of order $$p^2$$ is abelian.

Sketch of proof: Fix a prime $$p$$ and such a group $$G$$. The center $$Z(G)$$ of $$G$$ must be nontrivial by the class equation, and hence have order $$p$$ or $$p^2$$. So $$G/Z(G)$$ has order $$p$$ or $$1$$. So $$G/Z(G)$$ is cyclic. So $$G$$ is abelian.

If $$p < q$$ are prime numbers, then there is a nonabelian group of order $$pq$$ if and only if $$q \equiv 1 \pmod{p}$$, and in this case there is only one such nonabelian group (up to isomorphism).

Sketch of proof: Fix primes $$p < q$$ and a group $$G$$ of order $$pq$$. Use Sylow theory to conclude that $$G$$ has a unique (and hence normal) subgroup $$H_q$$ of order $$q$$, and a subgroup $$H_p$$ of order $$p$$, and use elementary observations to verify that $$H_q \cap H_p = \{1\}$$ and $$G = H_q H_p$$ so that $$G$$ is some semidirect product $$H_q \rtimes H_p$$. Since $$H_q$$ is cyclic, $$|\operatorname{Aut}(H_q)| = q-1$$, and so when $$q \not\equiv 1 \pmod{p}$$, Lagrange's theorem shows that there are no nontrivial homomorphisms $$H_p \to \operatorname{Aut}(H_q)$$, implying that $$G$$ is abelian in this case. When $$q = 1 \pmod{p}$$, the existence of a nontrivial homomorphisms $$H_p \to \operatorname{Aut}(H_q)$$ and the independence of the structure of the resulting semidirect product on the choice of nontrivial homomorphism are discussed in this math.SE post.

Together, these facts imply:

• There are no nonabelian groups of odd order less than $$21$$.

(The only odd numbers less than $$21$$ besides $$1$$ and primes to worry about are $$9$$ and $$15$$, which are covered by the above general results.)

• There is a unique nonabelian group of order $$21$$, and it can be realized by choosing any nontrivial homomorphism $$\mathbb{Z}/3\mathbb{Z} \to \operatorname{Aut}(\mathbb{Z}/7\mathbb{Z})$$ and forming the corresponding semidirect product. For example (taken from the above math.SE post), one could use the unique homomorphism sending $$1$$ to "multiplication by $$2$$".)

Now for my slightly subjective question:

Are there "less technical" or "more insightful" constructions of the nonabelian group of order $$21$$?

To be clearer about my kind of fuzzy requirements:

• I don't care if a construction doesn't shed any light on the nonexistence of nonabelian groups of smaller order.

• I do want it to be possible to define the group, and see that it is nonabelian, and of order 21, without drawing on too much group theoretical machinery.

• Ideally, the construction should minimize computation. One could, for example, exhibit two permutations of a small finite set that generate the group--- but verifying that the group they generate has 21 elements and is nonabelian is entirely computational. And even if the computations are well organized, this approach gives little idea of where such a group might "come from". (If a specific presentation came with a geometrical insight making clear why the resulting group would have order $$21$$ without computation, that would be awesome.)

To explain the motivation behind my question: I was teaching algebra, and was asked in office hours if nonabelian groups of odd order even existed. "Of course they do," I said. Then I realized that it was rather difficult to exhibit one, given where we were in the class. We had the basic homomorphism theorems, but:

• We hadn't done the Sylow theorems or semidirect products.
• We didn't know about finite fields yet.
• All of the members of the "families of groups" we had discussed up to that point--- symmetric groups, alternating groups, dihedral groups--- when nonabelian, have even order.
• All nonabelian groups of order less than $$21$$, whether or not they into the families just mentioned, have even order.

FWIW: the "most elementary" example of a small nonabelian group of odd order that I can think of is the set of all $$3 \times 3$$ matrices with entries in the field with $$3$$ elements that are both (a) upper triangular and (b) have all $$1$$s down the diagonal. (This is pretty clearly a nonabelian group, and just as clearly has $$27$$ elements.) This happens to be the next smallest nonabelian group of odd order.

• So essentially, you're asking for a more tangible construction of the odd group of order $21$? – Alex Becker Dec 22 '12 at 22:28

## 4 Answers

If $F$ is any field, there is a subgroup of $\text{PGL}_2(F)$ given by fractional linear transformation of the form $z \mapsto az + b$ where $a \in F^{\times}, b \in F$. This is precisely the subgroup fixing the point at infinity in $\mathbb{P}^1(F)$. Taking $F$ to be a finite field $\mathbb{F}_q$ we obtain a family of (usually) nonabelian Frobenius groups of order $q(q - 1)$.

Now, we can further restrict $a$ to lie in any subgroup of $F^{\times}$; in particular, for $q$ odd, taking it to lie in the subgroup of quadratic residues gives a family of (usually) nonabelian groups of order $\frac{q(q-1)}{2}$. Taking $q = 7$ gives the desired group.

Well, without Sylow theorems and semidirect products not many options remain open... and thus constructing such a group will look a little (or a lot) like out of the blue.

Anyway, the non-abelian group of order $$\,21\,$$ can be given as

$$G=\langle\,x,y\;\;;\;\;x^3=y^7=1\;,\;xy=y^2x\,\rangle\cong\langle\,(235)(476)\;,\;(1234567)\,\rangle\leq S_7$$

It is, I suppose, easier to show that the right hand group, with permutations, gives us what we want, but none is as easy as expressing the group as a semidirect product.

• Very sorry to have neglected to upvote this until yesterday when I rediscovered this thread. It may be worth pointing out that if $H$ is the subgroup generated by the 7-cycle $y$, and $K$ the subgroup generated by $x$, then the relation $xy=y^2x$ implies that $K\le N_{S_7}(H)$; a condition known to imply that $HK$ is a subgroup. Therefore $G=HK$ has exactly $21$ elements. Of course, it is easy to reach the same conclusion by directly proving that $HK$ is closed under products. Not much to this. – Jyrki Lahtonen Mar 6 '19 at 5:41

How about a mix of DonAntonio's two answers? He writes:

Anyway, the non-abelian group of order 21 can be given as $G=\langle\,x,y\;\;;\;\;x^3=y^7=1\;,\;xy=y^2x\,\rangle\cong\langle\,(235)(476)\;,\;(1234567)\,\rangle\leq S_7$

Give both the presentation with generators and relations, and its realization with x and y in $S_7$. It's easier to calculate with the former, and clear that the group has no more than the 21 elements $x^i y^j$ for $0 \leq i \leq 2$ and $0 \leq j \leq 6$. But it's not clear that the presentation doesn't "collapse" by having the given relations produce additional relations that identify some of these 21 elements. On the other hand, it's easy to show that the two permutations satisfy the relations, and the group they generate must have at least 21 elements by Lagrange's Theorem.

The group is the 'odd subgroup' in $GL_3(2)$.

OP specified the students haven't done finite fields, but if they've done regular matrices, then "vectors of bits" and "matrices of bits" may seem quite natural and certainly does not require explanation of finite fields in general.

The group clearly has 168 elements. If they've done Lagrange perhaps they might suspect there's a subgroup of order 21.

$$x=\begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{bmatrix}, y=\begin{bmatrix} 0 & 0 & 1 \\ 0 & 1 & 1 \\ 1 & 1 & 1 \end{bmatrix}$$ $x$ manifestly has order 3, and conjugating by $x$ just rotates the rows and columns of a matrix (this is true in any field). $y$ is not so manifest, but you can start from any non-zero vector and write out its images under repeated action of $y$, to show that it cycles through all seven non-zero vectors. This also lets you read off $y^2$. Hooray, $x^3 = 1, y^7 = 1, y^2 = y^x$.

• This is actually a nice idea. Unless I'm mistaken, powers of $y$ form the multiplicative group of a copy of the field $\Bbb{F}_8$ (so apart from occasionally summing to zero the set would be closed under addition as well). Squaring is an automorphism of that field. And by Skolem-Noether Theorem squaring can be achieved by conjugating with a suitable matrix. Squaring is an order three automorphism, so by adjusting that matrix (if necessary) we can make it order three (need multiplicative Hilbert 90 at that point). – Jyrki Lahtonen Mar 5 '19 at 19:58