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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.

Thanks in advance.

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  • $\begingroup$ Take a look at the semidirect product construction. $\endgroup$ – anomaly Feb 3 '17 at 23:35
  • $\begingroup$ See the section Generalized quaternion group of this page. $\endgroup$ – user 170039 Feb 4 '17 at 3:42
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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.

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  • $\begingroup$ Why is $S_3$ a dihedral group? $\endgroup$ – Pagginelli Feb 3 '17 at 23:56
  • $\begingroup$ 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. $\endgroup$ – Hans Musgrave Feb 4 '17 at 0:01
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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 $H\rtimes ({\bf Z}/2{\bf Z})$.

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"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 $e,\psi,\psi^2,\phi,\phi\psi,\phi\psi^2$.

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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.

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