Dihedral group $D_{8}$ as a semidirect product $V\rtimes C_2$? 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?
 A: 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.
A: If you know a bit more about group theory you could always do the following: 


*

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

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

*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 $D_8$.
A: 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 $X$;   $\phi(x)=\phi_x\in \text{Aut}(K)$ 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.
A: 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 $\pi\circ\phi=\mathrm{id}_{C_2}$.
