# Prob. 18, Sec. 2.3, in I.N. Herstein's TOPICS IN ALGEBRA, 2nd ed: For any $n > 2$ construct a non-abelian group of order $2n$

Here is Prob. 18, Sec. 2.3, in the book Topics in Algebra by I.N. Herstein, 2nd edition:

For any $$n > 2$$ construct a non-abelian group of order $$2n$$.

Herstein gives the following hint to this question: "imitate the relations in $$S_3$$ (permutation group of order $$3$$)".

I've already seen an answer to this problem using dihedral groups, but I still couldn't solve the question based on the hint.

Despite the fact I'm asking for a particular solution, please feel free to share different ways of doing it. It will be very interesting.

• Take a look at the semidirect product construction. Feb 3, 2017 at 23:35
– user170039
Feb 4, 2017 at 3:42

The hint was intended to point you toward dihedral groups, since $S_3$ is also a dihedral group of order $6$. By looking at the relations between elements and considering applying those same kinds of simple relations (half the elements are of order $2$, $r\cdot s^{-1}=s\cdot r$, etc...) to larger groups of order $2n$, you can uncover the dihedral groups.

• Why is $S_3$ a dihedral group? Feb 3, 2017 at 23:56
• It is isomorphic to $D_3$ (or $D_6$ depending on your notation), where the identity permutation maps to the identity in $D_3$, the transpositions map to reflections, and the two order-three elements map to the other two rotations in $D_3$. I don't think all such functions are mappings are isomorphisms necessarily (I haven't considered it in too much depth), but there is at least one. Feb 4, 2017 at 0:01

Hint: If $H$ is a cyclic group of order $n>2$, then $h\mapsto h^{-1}$ is a nontrivial automorphism (more generally, this is true if $H$ is abelian and not of exponent $2$). Deduce that there is a nonabelian semidirect product $H\rtimes ({\bf Z}/2{\bf Z})$.

First, refer to Example 2.2.3 in the book Topics in Algebra by I.N. Herstein, 2nd edition.

Imitating the relations in $$S_3$$, let us consider two objects $$\phi$$ and $$\psi$$ such that $$\phi^2 = \psi^n = e, \tag{0}$$ and $$\phi \psi = \psi^{n-1} \phi,$$ $$\phi \psi^2 = \psi^{n-2} \phi,$$ and so on $$\phi \psi^{n-1} = \psi \phi.$$

Along the lines of Example 2.2.3 in Herstein, let $$X$$ be the following set: $$X \colon= \left\{ \, x_1, \ldots, x_n \, \right\}.$$

Let us first suppose that our $$n$$ is an odd natural number greater than $$2$$.

And, let $$\phi$$ and $$\psi$$ be the following permutations of elements (i.e. bijective self-maps) of $$X$$: $$\phi \colon \begin{matrix} & x_1 \longrightarrow x_{n-1} \\ & x_2 \longrightarrow x_{n-2} \\ & x_3 \longrightarrow x_{n-3} \\ & \vdots \\ & x_{(n-1)/2} \longrightarrow x_{(n+1)/2} \\ & x_{(n+1)/2} \longrightarrow x_{(n-1)/2} \\ & \vdots \\ & x_{n-3} \longrightarrow x_3 \\ & x_{n-2} \longrightarrow x_2 \\ & x_{n-1} \longrightarrow x_1 \\ & x_n \longrightarrow x_n \end{matrix} \qquad \mbox{ in other words } \qquad x_i \longrightarrow \begin{cases} x_{n-i} \ & \mbox{ for } i = 1, 2, \ldots, n-1; \\ x_n \ & \mbox{ for } i = n; \end{cases}$$ and $$\psi \colon \begin{matrix} & x_1 \longrightarrow x_2 \\ & x_2 \longrightarrow x_3 \\ & x_3 \longrightarrow x_4 \\ & \vdots \\ & x_{n-1} \longrightarrow x_n \\ & x_n \longrightarrow x_1 \end{matrix} \qquad \mbox{ in other words } \qquad x_i \longrightarrow \begin{cases} x_{i+1} \ & \mbox{ for } i = 1, 2, \ldots, n-1; \\ x_1 \ & \mbox{ for } i = n. \end{cases}$$ Note that our $$n$$ here is a natural number (in fact an odd natural number) greater than $$2$$.

Then we note that $$\phi \psi \colon \begin{matrix} & x_1 \longrightarrow x_n \\ & x_2 \longrightarrow x_{n-1} \\ & x_3 \longrightarrow x_{n-2} \\ & x_4 \longrightarrow x_{n-3} \\ & \vdots \\ & x_{(n-1)/2} \longrightarrow x_{(n+3)/2} \\ & x_{(n+1)/2} \longrightarrow x_{(n+1)/2} \\ & \vdots \\ & x_{n-3} \longrightarrow x_4 \\ & x_{n-2} \longrightarrow x_3 \\ & x_{n-1} \longrightarrow x_2 \\ & x_n \longrightarrow x_1 \end{matrix} \qquad \mbox{ in other words } \qquad x_i \longrightarrow x_{n+1-i} \ \mbox{ for } i = 1, 2, \ldots, n,$$ whereas $$\psi \phi \colon \begin{matrix} & x_1 \longrightarrow x_{n-2} \\ & x_2 \longrightarrow x_{n-3} \\ & x_3 \longrightarrow x_{n-4} \\ & \vdots \\ & x_{(n-1)/2} \longrightarrow x_{(n-1)/2} \\ & x_{(n+1)/2} \longrightarrow x_{(n-3)/2} \\ & \vdots \\ & x_{n-2} \longrightarrow x_1 \\ & x_{n-1} \longrightarrow x_n \\ & x_n \longrightarrow x_{n-1} \end{matrix} \qquad \mbox{ in other words } \qquad x_i \longrightarrow \begin{cases} x_{n-1-i} \ & \mbox{ for } i = 1, 2, \ldots, n-2; \\ x_n \ & \mbox{ for } i = n-1; \\ x_{n-1} \ & \mbox{ for } i = n. \end{cases}$$ Thus we see that $$\phi \psi \neq \psi \phi. \tag{1}$$

Further we note that \begin{align} \psi^2 & \colon \begin{matrix} & x_1 \longrightarrow x_3 \\ & x_2 \longrightarrow x_4 \\ & x_3 \longrightarrow x_5 \\ & \vdots \\ & x_{n-2} \longrightarrow x_n \\ & x_{n-1} \longrightarrow x_1 \\ & x_n \longrightarrow x_2 \end{matrix} \qquad \mbox{ in other words} \qquad x_i \longrightarrow \begin{cases} x_{i+2} \ & \mbox{ for } i = 1, 2, \ldots, n-2; \\ x_1 \ & \mbox{ for } i = n-1; \\ x_2 \ & \mbox{ for } i = n; \end{cases} \\ \\ \psi^3 & \colon \begin{matrix} & x_1 \longrightarrow x_4 \\ & x_2 \longrightarrow x_5 \\ & \vdots \\ & x_{n-3} \longrightarrow x_n \\ & x_{n-2} \longrightarrow x_1 \\ & x_{n-1} \longrightarrow x_2 \\ & x_n \longrightarrow x_3 \end{matrix} \qquad \mbox{ in other words} \qquad x_i \longrightarrow \begin{cases} x_{i+3} \ & \mbox{ for } i = 1, 2, \ldots, n-3; \\ x_{n+1-i} \ & \mbox{ for } i = n-2, n-1; \\ x_3 \ & \mbox{ for } i = n; \end{cases} \\ \\ & \vdots \\ \\ \psi^{n-1} & \colon \begin{matrix} & x_1 \longrightarrow x_n \\ & x_2 \longrightarrow x_1 \\ & x_3 \longrightarrow x_2 \\ & x_4 \longrightarrow x_3 \\ & \vdots \\ & x_{n-1} \longrightarrow x_{n-2} \\ & x_n \longrightarrow x_{n-1} \end{matrix} \qquad \mbox{ in other words } \qquad x_i \longrightarrow \begin{cases} x_n \ & \mbox{ for } i = 1; \\ x_{i-1} \ & \mbox{ for } i = 2, 3, \ldots, n. \end{cases} \end{align}

In short, we note that, for each $$j = 1, 2, 3, \ldots, n-1$$, we have $$\psi^j \colon \qquad x_i \longrightarrow \begin{cases} x_{i+j} & \mbox{ if } i+j \leq n, \\ x_{i+j-n} & \mbox{ otherwise}. \end{cases} \ \mbox{ for each } i = 1, 2, \ldots, n. \tag{2}$$ And, of course $$\psi^n = e,$$ where $$e$$ denotes the identity element in $$S_n$$ which leaves every element of our set $$X$$ fixed.

Now we see that $$\psi^{n-1} \phi \colon \begin{matrix} & x_1 \longrightarrow x_n \\ & x_2 \longrightarrow x_{n-1} \\ & x_3 \longrightarrow x_{n-2} \\ & x_4 \longrightarrow x_{n-3} \\ & \vdots \\ & x_{n-1} \longrightarrow x_2 \\ & x_n \longrightarrow x_1 \end{matrix} \qquad \mbox{ in other words } \qquad x_i \longrightarrow x_{n+1-i} \ \mbox{ for } i = 1, 2, \ldots, n;$$ this shows that $$\psi^{n-1} \phi = \phi \psi.$$

In fact, for each $$j = 1, 2, \ldots, n-1$$, we see that $$\phi \psi^j \colon \qquad x_i \longrightarrow \begin{cases} x_{n-i+j} \ & \mbox{ for } i = 1, 2, \ldots, n-1 \mbox{ such that } i \geq j; \\ x_{j-i} \ & \mbox{ for } i = 1, 2, \ldots, n-1 \mbox{ such that } i < j; \\ x_j \ & \mbox{ for } i = n. \end{cases} \tag{3}$$

On the other hand, for each $$j = 1, 2, \ldots, n-1$$, we see that
$$\psi^{n-j} \colon \qquad x_i \longrightarrow \begin{cases} x_{n + i -j} & \mbox{ if } i \leq j , \\ x_{i-j} & \mbox{ if } i > j \end{cases} \ \mbox{ for each } i = 1, 2, \ldots, n.$$ Therefore, for each $$j = 1, 2, \ldots, n-1$$, we have $$\psi^{n-j} \phi \colon \qquad x_i \longrightarrow \begin{cases} x_{j-i} \ & \mbox{ for } i = 1, 2, \ldots, n-1 \mbox{ such that } i < j; \\ x_{n-i+j} \ & \mbox{ for } i = 1, 2, \ldots, n \mbox{ such that } i > j; \\ x_n \ & \mbox{ for } i = 1, 2, \ldots, n-1 \mbox{ such that } i = j. \end{cases} \tag{4}$$

From (3) and (4) we get $$\phi \psi^j = \psi^{n-j} \phi. \tag{5}$$ for each $$j = 1, 2, \ldots, n-1$$. And, this equality holds of course for $$j = 0$$ or $$j = n$$.

Hence for our desired non-abelian group of order $$2n$$ we have the set $$G \colon= \left\{ \, e, \psi, \psi^2, \ldots, \psi^{n-1}, \phi, \phi \psi, \phi \psi^2, \ldots, \phi \psi^{n-1} \, \right\}$$ under the binary operation (of composition of mappings) such that $$\psi^n = e = \phi^2;$$ $$\phi \psi^j = \psi^{n-j} \phi \ \mbox{ for each } j = 0, 1, 2, \ldots, n-1, n;$$ and of course $$\phi \psi \neq \psi \phi$$ because here our $$n > 2$$.

I'll discuss the case of an even natural number $$n > 2$$ later.

"Imitate the relations in $S_3$": To get the context of that hint, see a few pages earlier in the chapter, Example 2.2.3, where there is elaboration on how $S_3$ is generated by $\phi$ and $\psi$ with orders $2$ and $3$ respectively such that $\phi\psi=\psi^{-1}\phi$. It is then explained that it follows that $e,\phi,\psi,\psi^2,\phi\psi$, and $\psi\phi$ are all of the elements of $S_3$. To imitate this, you can change $3$ to $n$ and see what happens. A more systematic way to organize the elements of $S_3$ that would make it easier to see how it will generalize may be $e,\psi,\psi^2,\phi,\phi\psi,\phi\psi^2$.

I am reading "Topics in Algebra 2nd Edition" by I. N. Herstein.
This problem is Problem 18 on p.36.
I solved this problem as follows:

Let $$\phi:=\begin{pmatrix}-1&0\\0&1\end{pmatrix}$$.
Let $$\psi:=\begin{pmatrix}\cos\frac{2\pi}{n}&-\sin\frac{2\pi}{n}\\\sin\frac{2\pi}{n}&\cos\frac{2\pi}{n}\end{pmatrix}$$.
Let $$D:=\{I_2,\psi,\psi^2,\dots,\psi^{n-1},\phi,\phi\cdot\psi,\phi\cdot\psi^2,\dots,\phi\cdot\psi^{n-1}\}$$.

Obviously, $$\psi^i\neq\psi^j$$ for $$0\leq i.
Obviously, $$\phi\psi^i\neq\phi\psi^j$$ for $$0\leq i.
Since $$\det\psi^i=1$$ and $$\det\phi\psi^j=-1$$, $$\psi^i\neq\phi\psi^j$$ for $$0\leq i.
So, $$\#D=2n$$.

$$\phi^2=I_2$$.

$$\psi^n=I_2$$.

$$\phi\cdot\psi=\begin{pmatrix}-1&0\\0&1\end{pmatrix}\cdot\begin{pmatrix}\cos\frac{2\pi}{n}&-\sin\frac{2\pi}{n}\\\sin\frac{2\pi}{n}&\cos\frac{2\pi}{n}\end{pmatrix}=\begin{pmatrix}-\cos\frac{2\pi}{n}&\sin\frac{2\pi}{n}\\\sin\frac{2\pi}{n}&\cos\frac{2\pi}{n}\end{pmatrix}$$.
$$\psi^{-1}\cdot\phi=\begin{pmatrix}\cos\frac{-2\pi}{n}&-\sin\frac{-2\pi}{n}\\\sin\frac{-2\pi}{n}&\cos\frac{-2\pi}{n}\end{pmatrix}\cdot\begin{pmatrix}-1&0\\0&1\end{pmatrix}=\begin{pmatrix}-\cos\frac{-2\pi}{n}&-\sin\frac{-2\pi}{n}\\-\sin\frac{-2\pi}{n}&\cos\frac{-2\pi}{n}\end{pmatrix}$$.
So, $$\phi\cdot\psi=\psi^{-1}\cdot\phi$$.

$$\phi\cdot\psi^{-1}=\phi\cdot\psi^{n-1}=(\phi\cdot\psi)\cdot\psi^{n-2}=\psi^{-1}\cdot(\phi\cdot\psi^{n-2})=\cdots=(\psi^{-1})^{n-1}\cdot\phi=\psi^n\cdot(\psi^{-1})^{n-1}\cdot\phi=\psi\cdot\phi.$$

$$\psi^i\cdot\phi\psi^j=\phi\psi^{-i}\psi^j=\phi\psi^{j-i}$$.
Since $$\psi^n=I_2$$, we can write $$\phi\psi^{j-i}=\phi\psi^k$$, where $$0\leq k.
So, $$\psi^i\cdot\phi\psi^j\in D$$.

$$\phi\psi^i\cdot\psi^j=\phi\psi^{i+j}$$.
Since $$\psi^n=I_2$$, we can write $$\phi\psi^{i+j}=\phi\psi^k$$, where $$0\leq k.
So, $$\phi\psi^i\cdot\psi^j\in D$$.

$$\psi^i\cdot\psi^j=\psi^{i+j}$$.
Since $$\psi^n=I_2$$, we can write $$\psi^i\cdot\psi^j=\psi^k$$, where $$0\leq k.

$$\phi\psi^i\cdot\phi\psi^j=\phi\phi\psi^{j-i}=\psi^{j-i}$$.
Since $$\psi^n=I_2$$, we can write $$\phi\psi^i\cdot\phi\psi^j=\psi^k$$, where $$0\leq k.

So, $$D$$ is closed under the matrix multiplication.

$$\psi^i\cdot\psi^{n-i}=I_2$$.
$$\phi\psi^i\cdot\phi\psi^i=I_2$$.
So, any element of $$D$$ has its inverse.

Therefore, $$D$$ is a subgroup of $$GL(2,\mathbb{R})$$.

$$\psi\cdot\phi=\phi\psi^{-1}=\phi\psi^{n-1}\neq\phi\psi$$.
So, $$D$$ is non-abelian.