The particular problem that I am facing is that when $H$ is a normal subgroup of order $7$ and $K$ is a subgroup of order $8$ .Now, we can consider the map $\theta :K \to Aut(H)$ we see that $\theta(k) \to khk^{-1}$ and $khk^{-1}=h^{-1}$.
I am unable to proceed after this .I need to prove that :
$a)$ when $K$ is isomorphic to $Z_2 \times Z_2 \times Z_2$ there is only one group which is non Abelian.
$(b) $ when $K = Z_4 \times Z_2$ there are two non isomorphic groups.
$(c) $ when $K=Z_8$ there is only one group
Best Answer
${\rm Aut}(H) \cong C_6$ is cyclic of order $6$. So in all cases, the non-trivial homomorphisms $\theta$ all have image of order $2$ and kernel $L$ of order $4$.
So you just need to find the normal subgroups $L$ of $K$ of order $4$, and then determine which of these give rise to isomorphic groups $H \ltimes K$.
In Case a), we have $L \cong C_2 \times C_2$, and all such $K$ give rise to isomorphic groups $H \ltimes K$ (note that all such $L$ are equivalent under ${\rm Aut}(K)$).
In Case b), we can have $L \cong C_4$ or $L \cong C_2^2$. The $C_2^2$ is unique. There are two copies of $C_4$, but they give isomorphic $H \ltimes K$.
In Case c) we must have $L=C_4$.