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How to construct a nonabelian group of order $n\phi (n)$, where $\phi (n)$ is the Euler phi function of $n$ for $n \geq 3$.

I am trying to use the fact that $\alpha : \mathbb{Z}_n^\times \to \operatorname{Aut}(\mathbb{Z}_n)$ is an isomorphism where $\mathbb{Z}_n^\times$ is the units of $\mathbb{Z}_n$. So we know that $|\mathbb{Z}_n^\times| = \phi (n)$, but I'm not sure where to go from here.

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    $\begingroup$ Do you know how to construct a group from an action of one group on another group? $\endgroup$ – Eric Wofsey Jul 21 '18 at 20:04
  • $\begingroup$ Looks like you may want the Holomorph of a group. $Hol(\Bbb{Z}_n)$ to be more precise. Groupprops has more about it. $\endgroup$ – Jyrki Lahtonen Jul 21 '18 at 20:21
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On a far more elementary level, why not take all linear polynomials $ax+b$, with $a,b\in\Bbb Z/n\Bbb Z$, under composition? Here of course $a$ must be invertible.

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    $\begingroup$ Or, what’s the same thing, all invertible matrices $\pmatrix{a&b\\0&1}$ over $\Bbb Z/n\Bbb Z$. $\endgroup$ – Lubin Jul 22 '18 at 13:26
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Hint:

Consider the semi-direct product $\;\mathbf Z_n\rtimes_f\mathbf Z_n^\times$.

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  • $\begingroup$ I see that this is nonabeilan but I'm not seeing why the order would be $n \phi (n)$ since we are not guaranteed that $\mathbb{Z}_n \cap \mathbb{Z}_n^\times = \{e\}$. $\endgroup$ – frostyfeet Jul 21 '18 at 21:55
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    $\begingroup$ As a set, it is just the direct product. $\endgroup$ – Bernard Jul 21 '18 at 22:06
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Pick any prime $p$ such that $p$ does not divide $n$ and $p\not\equiv 1\pmod{n}$. Such a $p$ exists as long as $n\geq 3$. Then $\sigma:\alpha\mapsto p\alpha$ is a nontrivial automorphism of the group $\mathbf{Z}/n\mathbf{Z}$, and the unique homomorphism $\mathbf{Z}\to\mathrm{Aut}(\mathbf{Z}/n\mathbf{Z})$ sending $1$ to $\sigma$ kills the subgroup $\varphi(n)\mathbf{Z}$ because $p^{\varphi(n)}\equiv 1\pmod{n}$. Therefore the homomorphism descends to a nontrivial homomorphism $\Sigma:\mathbf{Z}/\varphi(n)\mathbf{Z}\to\mathrm{Aut}(\mathbf{Z}/n\mathbf{Z})$ satisfying $\Sigma(1+\varphi(n)\mathbf{Z})=\sigma$.

The semidirect product $\mathbf{Z}/n\mathbf{Z}\rtimes_\Sigma\mathbf{Z}/\varphi(n)\mathbf{Z}$ with respect to the homomorphism $\Sigma$ will then be a nonabelian group of order $n\varphi(n)$.

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The affine group over $\Bbb{Z}/n\Bbb{Z}$ has order $n\phi(n)$, and is the group Lubin described. We have $$ {\rm Aff}(\Bbb{Z}/n\Bbb{Z})\cong \Bbb{Z}/n\Bbb{Z}\rtimes U(\Bbb{Z}/n\Bbb{Z}). $$

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