Construct group $G=Q\ltimes Z_{7}$ I read a example. i have problem.
Let $Q=\langle x, y| x^{9}=y^{3}=1, x^{y}=x^{4}\rangle$ and $Z_{7}$ be a group of order $7$.
I know $\Omega_{1}(Q):=\Omega$ is an abelian group of order $9$. Since 
$Q/\Omega$ is isomorphic to the subgroup $Z_{3}$ of order $3$ of the automorphism group $Aut(Z_{7})$. we may form $G=Q\ltimes Z_{7}$.
Do you need $\Omega_{1}$ for construct G ?
$\Omega_{i}=$ the subgroup of $G$ generated by its elements of order dividing $p^{i}$ 
 A: If you use $\Omega_1$ this way you will miss some possibilities for $G=Q\ltimes Z_7$. The way you describe gives one of the four possibilities. So if you want to specify that one possibility, then yes, use $\Omega_1$. If you want all possibilities, then no, do not use it.
Since the Sylow 3-subgroup of $\operatorname{Aut}(Z_7)$ has order 3, you know that $x^3$ acts as the identity on $Z_7$ for all $x \in Q$. Hence you get an action of $Q/\mho(Q)$ where $\mho_1(Q) = \mho(Q) = \langle x^3 : x \in Q \rangle$.  However, $G/\mho(Q) \cong C_3 \times C_3$ has many homomorphisms to $\operatorname{Aut}(Z_7)$.
In particular, there are four distinct semidirect products $G$ (one of them direct). You can specify one of them by requiring that $\Omega(Q)$ is the kernel of $Q \to \operatorname{Aut}(Z_7)$.
$Q$ has 2 $\operatorname{Aut}(Q)$-orbits of subgroups of order 9. One is $\{ \Omega_1(Q) \}$ and the other, $\{ \langle x \rangle, \langle yx \rangle, \langle y^2x \rangle \}$, consists of 3 cyclic subgroups of order 9. The semidirect products with kernel $\Omega_1(Q)$ are all isomorphic, but if the kernel is in the other orbit, then there are two possible actions, $z^y = z^2$ or $z^y = z^4$, giving two more semidirect products. Also the kernel could be order 27, giving the direct product as a fourth possibility.
