Krdytkyu

How do I show that the dihedral group $D_{8}$ (order $8$) is a semidirect product $Vrtimes leftlangle alpha rightrangle $, where $V$ is Klein group and $%
alpha $ is an automorphism of order two?

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 $D_8$.

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

One way would be to show that if you take $V=C_2times C_2$, $langle alpha rangle$ injects into the automorphism group of $V$, which is isomorphic to $GL_2(mathbb{F}_2)$, by mapping $alpha$ to $$left(begin{array}{cc}1&1\0&1end{array}right).$$
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 $$langle r,s,t|r^2=s^2=t^2=[r,s]=[s,t]=1,r^t=rsrangle.$$

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

I think the following fact can help us:

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

Now put $H=langle xmid x^p=1 rangle$ and $K=langle ymid y^{p^{m-1}}=1 rangle$ wherein $p$ is a prime and $minmathbb N$. We can consider $forall xin H, phi(x)=phi_x$ such that $phi_x(y)=x^{-1}yx, yin K$. 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: $$M_m(p)=langle x,ymid x^p=1,y^{p^{m-1}}=1,x^{-1}yx=y^{p^{m-2}}rangle$$

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

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 $Nsubset G$. Denote by $pi:Gto G/N$ the projection map. Then $Gcong Nrtimes G/N$ iff there exists a homomorphism $phi: G/Nto G$, such that $picircphi=mathrm{id}_{G/N}$. This $phi$ is called a splitting homomorphism.

Back to our dihedral group: $D_8/Vcong C_2$. In order to define a splitting homomorphism $C_2to D_8$ 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 $0in C_2$ because it does not lie in $V$. So we constructed a homomorphism $phi: C_2to D_8$ such that $picircphi=mathrm{id}_{C_2}$.