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How do I show that the dihedral group $D_{8}$ (order $8$) is a semidirect product $V\rtimes \left\langle \alpha \right\rangle $, where $V$ is Klein group and $% \alpha $ is an automorphism of order two?

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4 Answers 4

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If you know a bit more about group theory you could always do the following:

  1. There is a non-trivial semi-direct product of $V$ and $C_2$ since $\mathrm{Aut}(V)$ contains an element of order $2$.

  2. The resulting group has order $8$ and is not abelian (we know that there is an element which acts non-trivial by conjugation).

  3. The resulting group has at least as many subgroups of order $2$ as $V$ so there are $3$ or more.

Finally: There is only one non-Abelian group of order $8$ with more than one subgroup of order $2$ and that is $D_8$.

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Lima: thank you! –  User2040 Nov 22 '12 at 23:03

Depends on what you're using as your definition of $D_8$.

One way would be to show that if you take $V=C_2\times C_2$, $\langle \alpha \rangle$ injects into the automorphism group of $V$, which is isomorphic to $GL_2(\mathbb{F}_2)$, by mapping $\alpha$ to $$\left(\begin{array}{cc}1&1\\0&1\end{array}\right).$$ Then show that the semidirect product defined by this automorphism is isomorphic to $D_8$ by whatever presentation you're using.

You could try the same idea using permutation group notation, taking $V=\langle(1,2),(3,4)\rangle$ and mapping $\alpha$ to the automorphism $$\varphi_\alpha:\left\{\begin{array}{l}(1,2)\mapsto (1,2)(3,4) \\ (3,4)\mapsto (3,4)\end{array}\right. .$$

Or yet another variation would be the presentation $$\langle r,s,t|r^2=s^2=t^2=[r,s]=[s,t]=1,r^t=rs\rangle.$$

It all depends how you want to do it, but this automorphism is the key.

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I see your points now here, Alexander, The Great. :) –  Babak S. Nov 22 '12 at 7:40

I think the following fact can help us:

Let $H$, $K$ be groups and $$\phi: H \overset{hom}{\longrightarrow}\text{Aut}(K)$$ be an automorphism of $K$. We denote for all $x$ in $X$; $\phi(x)=\phi_x\in \text{Aut}(K)$ Then the product $$K\times H$$ with the following rule can make it a new structure called the semi-product of $K$ by $H$ and denote that by $$K\times_{\phi}H\\(k,h)\cdot(k',h')=(k\phi_{h}(k'),hh')$$ In fact, the homomorphism $\phi$ caused a rule for group operation.

Now put $H=\langle x\mid x^p=1 \rangle$ and $K=\langle y\mid y^{p^{m-1}=1} \rangle$ wherein $p$ is a prime and $m\in\mathbb N$. We can consider $\forall x\in H, \phi(x)=\phi_x$ such that $\phi_x(y)=x^{-1}yx, y\in K$. Here, we have $$\phi_{x}: y \mapsto y^{p^{m-2}}$$ and it is really of order $p$. now by appling the above nwe structure we have: $$M_m(p)=\langle x,y\mid x^p=1,y^{p^{m-1}}=1,x^{-1}yx=y^{p^{m-2}}\rangle$$

It is easily be seen that $M_3(2)$ has the same presentation as $D_8$ has.

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ops! The group K the Klein would be?? –  User2040 Nov 22 '12 at 0:13
    
@Lima: Actually, no. In fact $D_{2n}$ is an extension of $\mathbb Z_n$ by $\mathbb Z_2$. So $K\cong\mathbb Z_4$ here. :) –  Babak S. Nov 22 '12 at 4:00
    
@BabakSorouh It's also an extension of $V$ by $\mathbb{Z}_2$. –  Alexander Gruber Nov 22 '12 at 7:15
    
@AlexanderGruber: Isn't it because the automorphism groups of $Z_4$ and $V$ are the same group? I, recently, worked on one of Rotman's book and what I thought about $D_8$ above came from there. –  Babak S. Nov 22 '12 at 7:38
    
Lima: thank you! –  User2040 Nov 22 '12 at 23:08

I like to think about $D_8$ as the group of invariants of a square.

So the our group $V$ is given by the identity, the 2 reflections that have no fix points and their product which is 180°-rotation.

Semidirect product can be characterized as split short exact sequences of groups. This is just a fancy way to say the following:

Given a group $G$ and a normal subgroup $N\subset G$. Denote by $\pi:G\to G/N$ the projection map. Then $G\cong N\rtimes G/N$ iff there exists a homomorphism $\phi: G/N\to G$, such that $\pi\circ\phi=\mathrm{id}_{G/N}$. This $\phi$ is called a splitting homomorphism.

Back to our dihedral group: $D_8/V\cong C_2$. In order to define a splitting homomorphism $C_2\to D_8$ we need to find an element of order 2 that is not contained in $V$. Such an element is given by a reflection through one of the diagonals of our square. It is clear that $\pi$ doesn't map t his element to $0\in C_2$ because it does not lie in $V$. So we constructed a homomorphism $\phi: C_2\to D_8$ such that $\pi\circ\phi=\mathrm{id}_{C_2}$.

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Lima: thank you! –  User2040 Nov 22 '12 at 23:03

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