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I just learnt semidirect product, but only know the basic definition, not gaining the true understanding of it. There is an example that asks the reader to construct a nonabelian group of order $21$. I want to learn this example to realize the semidirect product. Need help.

Let $H=\Bbb Z_7,~K=\Bbb Z_3, \phi:\Bbb Z_3\to\text{Aut}(\Bbb Z_7)\cong\Bbb Z_6$. Then I know that $H$ and $K$ can make a semidirect product. However, what $\phi$ should I choose (can I choose the trivial homomorphism? or something else?), and why is the resulting $G$ is $G:=\left\langle x,y | x^7=e=y^3, y^{-1}xy=x^2\right\rangle$? Where is "$y^{-1}xy=x^2$" come from?

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Here $y^{-1}xy=x^2$ is making conjugation by $y$ act as an automorphism of the group generated by $x$ - it sends $x$ to another generator, namely $x^2$.

Note it is important that this automorphism has order $3$ so

$$\begin{align} y^{-3}xy^3 &=y^{-2}x^2y^2 \\ &=y^{-1}x^4y \\ &=x^8\\ &=x \end{align}$$

calculating by iterating the automorphism, but also we have $y^3=1$, so the two are compatible.

These are the components you need to make the semidirect product construction work.

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The trivial homomorphism yields the direct product, so you should not use that one. But any other homomorphism yields a non-trivial semidirect product. There are two nontrivial homomorphisms $\varphi: (\mathbb{Z}_3,+)\rightarrow (\mathbb{Z}_6,+)$. It doesn't matter which one you pick, the results will be isomorphic groups. I pick $\varphi: (\mathbb{Z}_3,+)\rightarrow (\mathbb{Z}_6,+)$, $x\mapsto 2x$. (The other option would be $4x$.) This homomorphism represents the action of the complement on the kernel by conjugation. The generator $1$ of $(\mathbb{Z}_3,+)$ is mapped to $2$ by $\varphi$ in $(\mathbb{Z}_6,+)$. The identification $(\mathbb{Z}_6,+)\cong Aut((\mathbb{Z}_7,+))\cong (\mathbb{Z}_7^\times,\cdot)$ is by finding the primitive root $3\in (\mathbb{Z}_7^\times,\cdot)$, and then mapping every element $n\in (\mathbb{Z}_6,+)$ to $3^n\in (\mathbb{Z}_7^\times,\cdot)$. In particular, the image of $2$ is $3^2\equiv 2$ in $(\mathbb{Z}_7^\times,\cdot)$. That is why conjugation by the generator of the complement is taking second powers, i.e., $y^{-1}xy=x^2$ for all $x\in (\mathbb{Z}_7,+)$, if $y$ is the generator of $(\mathbb{Z}_3,+)$.

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  • $\begingroup$ Thanks. I stuck in "the action of the complement on the kernel by conjugation". What does it mean formally? Is it part of a definition in semidirect product? $\endgroup$ – Eric Dec 31 '18 at 10:41
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    $\begingroup$ II think you should first read up on semidirect products from a textbook, and then check my answer again. $\endgroup$ – A. Pongrácz Dec 31 '18 at 10:42
  • $\begingroup$ @A.Pongrácz You can also try to see if you can clarify that sentence, perhaps? It's not too much work. I don't understand what you mean by that either. $\endgroup$ – Pedro Tamaroff Dec 31 '18 at 13:41
  • $\begingroup$ @Pedro Tamaroff If I were lazy to help, I would not have written up the answer at the first place. The reason I am not doing that is what I wrote: I think that the author of the post should put more of his or her own effort into understanding the basics. Spoonfeeding is not helpful. $\endgroup$ – A. Pongrácz Dec 31 '18 at 13:44
  • $\begingroup$ @A.Pongrácz There is a difference between not understanding a concept or definition and not understanding the way these are being explained to you. The problem here is how you have phrased your sentence, which seems to be a bit problematic for the OP, and perhaps not a lack of reading from the OP's part. $\endgroup$ – Pedro Tamaroff Dec 31 '18 at 13:46

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