# Understanding semidirect product by constructing a non-abelian group of order $21$

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?

## 2 Answers

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.

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 $$(\mathbb{Z}_3,+)$$.

• 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? – Eric Dec 31 '18 at 10:41
• II think you should first read up on semidirect products from a textbook, and then check my answer again. – A. Pongrácz Dec 31 '18 at 10:42
• @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. – Pedro Tamaroff Dec 31 '18 at 13:41
• @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. – A. Pongrácz Dec 31 '18 at 13:44
• @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. – Pedro Tamaroff Dec 31 '18 at 13:46