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I'm studying for the first time semidirect product and I'm trying to learn how to identify some of them.

For example $\mathbb{Z}/4\mathbb{Z} \rtimes \mathbb{Z}/2\mathbb{Z}$

I red that, for $H\rtimes K$ consists to identify all the homomorphisms between $K$ and $Aut(H)$ but I don't understand why. The only thing we've seen at class is the definition of a semi-direct product with the actions.

Maybe it's a stupid question. In this example there is only one homomorphism but I want to understand why I'm doing this.

Thank you.

MaríaCC
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  • What was the definiton of semi-direct product that you were given in class? – hmmmm Feb 27 '14 at 16:48
  • We have $C$ acting on $A$ by the homomorphism $\tau:C\longrightarrow S_A$. With the composition: $$(a,c),(a',c')\mapsto (a\tau(c)(a'),cc')$$, $A\times C$ it's a group named semidirect group of $C$ by $A$ related to $\tau$. – MaríaCC Feb 27 '14 at 16:55
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    That group is the dihedral group $;D_8;$...can you prove that it can't be the quaternion group $;Q_8;$ ? – DonAntonio Feb 27 '14 at 18:40
  • I don't understandthe question? Why I'm supposed to prove that? – MaríaCC Feb 27 '14 at 19:28
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    @MaríaCC, because if you want a non-trivial semi direct product (i.e., not a direct product) then this one must be non-abelian, and there are only two non-abelian groups of order eight... – DonAntonio Feb 27 '14 at 20:27

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What I know about this group is: Let $N=\mathbb Z_4=\langle x\rangle$ and $H=\mathbb Z_2=\langle y\rangle$ so, $x^4=1$ and $y=y^{-1}$. We know that $Aut(N)\cong H$ and so as you pointed, there is a homomorphism $$\phi:H\to Aut(N)\\\phi_y(x)=x^l,~~~~~ 0\leq l\leq3$$ Since $\phi_y(x)$ is an automorphism and $|N|=4$ so we should have $\text{gcd}(l,4)=1$. But $y^2=1$ makes us to have $$(\phi_y)^2(x)=x^{l^2}=1$$ So $l^2\equiv 1~~(\text{mod} ~~4)$ so $l=1$ or $l=3$. If $l=1$ your group is abelian but this group as @Don's indicated is non-abelian so $l=3$ and therefore: $$x^4=1,y^2=1,y^{-1}xy=x^3\to x^4=1,y^2=1,yxy=x^{-1}\to x^4=1,y^2=1,(yx)^2=1$$ which presented the group $D_8$.

Mikasa
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