Non-abelian group of order $2n$

I am in the begining of studying abstract algebra and I ran into the following problem from Herstein's book:

For any $n>2$ construct a non-abelian group of order $2n$. (Hint: imitate the relations in $S_3$.)

My efforts: I was trying to realize the meaning of hint but my efforts were unsuccesful. We know that $S_3$ is non abelian group and $S_3=\{e,\ \phi,\ \psi,\ \phi\circ \psi,\ \psi\circ \phi, \ \psi^2\},$ where $$\phi:\{x_1,x_2,x_3\}\to \{x_2,x_1,x_3\} \quad \text{and} \quad \phi:\{x_1,x_2,x_3\}\to \{x_2,x_3,x_1\}$$

But I do not know how to apply the structure of $S_3$ with order $3!=6$ to the group $G$ with order $2n$ .

I did not find anything useful in this site.

• Here's a (hint at a) way to answer your question geometrically: study the symmetries of a regular $n$-gon. For $n=3$ you get $S_3$. – Ethan Bolker Nov 22 '17 at 14:59
• The symmetries of an equilateral triangle are rotations and the reflections over the three axes of symmetry.. Write them down and write down the multiplication table. Look at the relations and convince yourself that the group is $S_3$. Now find and study the $8$ symmetries of the square. I'm not prepared to "show a detailed answer" for you. Maybe someone else will, but then you won't have the fun of figuring it out for yourself. – Ethan Bolker Nov 22 '17 at 15:06
• @RFZ the reason why Ethan Bolker is considering symmetry is that is what groups are for. The primary role of groups in mathematics is to represent the symmetries of something. – fredgoodman Nov 22 '17 at 15:20
• You may want to look up Dihedral groups. These groups are essentially the symmetries Ethan Bolker mentioned. – edm Nov 22 '17 at 15:32
• If by Herstein's book you mean Topics in Algebra, be aware that many of his exercises are hard to do with the limited information available up to that point in the book, but they become simple once you have a bit more knowledge. – Bungo Nov 22 '17 at 15:36

Here is what I take Herstein's hint to mean.

$S_3$ is generated by elements $\phi$ and $\psi$ satisfying the relations $\phi^2=1$, $\psi^3=1$ and $\psi\phi = \phi\psi^{-1}$. Herstein is quoted as writing: Hint: imitate the relations in $S_3$. This suggests that he intends you to examine a possible group generated by elements $\Phi$ and $\Psi$ satisfying $\Phi^2=1$, $\Psi^n=1$ and $\Psi\Phi = \Phi\Psi^{-1}$. Is there such a group? Does it have order $2n$? Is it noncommutative?

You already have clues to the answer, because in $S_3$, the group of permutations of $\{1, 2, 3\}$, you indicated that you know that the permutations $\phi = (1\;2)$ and and $\psi = (1\;2\;3)$ generate the group and (it can be checked that they) satisfy the relations. This might suggest to you that you try to check whether the permutations $\Phi, \Psi$ of $\{1,\ldots,n\}$ defined by $\Phi=(1\;2)$ and $\Psi=(1\;2\;\cdots\;n)$ generate a group satisfying the desired conditions.

It turns out that they do not satisfy the relations. Here is where the comments to the question come in. It is better to choose $\Phi=(1\;n)(2\;n-1)\cdots(\lfloor\frac{n}{2}\rfloor\;\lceil\frac{n}{2}\rceil)$ and $\Psi=(1\;2\;\cdots\;n)$. This choice looks a little weird, but if you label the vertices of a regular $n$-gon with the numbers $1, 2, \cdots, n$, then $\Psi$ represents a $2\pi/n$ rotation and $\Phi$ represents a reflection through some axis of symmetry.

Could you demonstrate entire and detailed sokution?

Since Fred Goodman already commented on your problem, let me direct you to his algebra book, which is free online. He writes about groups satisfying $\Phi^2=1$, $\Psi^n=1$ and $\Psi\Phi = \Phi\Psi^{-1}$ in Section 2.3.

• Could you demonstrate entire and detailed sokution? It would be great – ZFR Nov 22 '17 at 16:28
• There are many entire and detailed solutions here on MSE. Start, say, from here. – Dietrich Burde Nov 22 '17 at 21:15