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I.N.Herstein - Topics in Algebra, 2nd edt.

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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    Take a look at the semidirect product construction. – 2017-02-03
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    See the section Generalized quaternion group of [this page](https://www.google.co.in/url?sa=t&rct=j&q=&esrc=s&source=web&cd=1&cad=rja&uact=8&ved=0ahUKEwieuvbkwvXRAhXMOI8KHYBYAYoQFggbMAA&url=https%3A%2F%2Fen.wikipedia.org%2Fwiki%2FQuaternion_group&usg=AFQjCNHCE0EL-qubdqpydOBjN9Iny5snxw). – 2017-02-04

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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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    Why is $S_3$ a dihedral group? – 2017-02-03
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    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. – 2017-02-04
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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 $e,\psi,\psi^2,\phi,\phi\psi,\phi\psi^2$.