prove that $D_8 \cong C_2 \wr C_2$ .
$\wr$ is wreath product and it is using in place of $C_2 \ltimes (C_2 \times C_2)$.
here is my answer :
suppose $K=C_2 \times C_2$ and $C_2 \cong \langle \sigma \rangle$ which $\sigma = (1,2)$.
consider homomorphism $\varphi : \langle \sigma \rangle \rightarrow Aut(k)$ which $$\sigma ^* :k \rightarrow k $$ $$((a,b) \rightarrow (a,b))$$
now if we use standard representation of $C_2 \times _{\varphi} (C_2 \times C_2) $we have :
$$C_2 \times _{\varphi} (C_2 \times C_2) \cong \langle x,y,z \mid x^2=y^2=z^2=1 ,xy=zy,x^{-1}yx=z,x^{-1}zx=y \rangle$$
with omitting generator $z$ we have :
$$C_2 \times _{\varphi} (C_2 \times C_2) \cong \langle x,y \mid x^2=y^2=(xy)^2=1 \rangle$$
I think my answer is right but I want to answer it without using standard representation, that I wasn't successful.
thank you very much.
$C_2 \ltimes (C_2 \times C_2)$ is a nonabelian group of order $8$, hence is either $D_8$ or $Q_8$ - but every subgroup of $Q_8$ is normal.