On the irreducible action Let $G=P \ltimes H$ be a non-nilpotent group, where $N_G(P)=P=Q_8$ and $H \cong \Bbb{Z}_q \times \Bbb{Z}_q$, ($q$ is a prime number). If $P$ acts irreducibly on $H$, then why the action is faithful?
I said that, suppose contrary that $C_P(H) \neq 1$. Thus $\langle x \rangle \leq Z(G)$, where $|x|=2$. I dont know how we get a contradiction. Also i seen this page https://groupprops.subwiki.org/wiki/Faithful_irreducible_representation_of_quaternion_group
in here, the irreducible peresentation of Q_8 is faithful, but i dont know why this holds.
 A: Edit: Because $G$ was assumed not to be nilpotent, we must have $q\neq2$. For the same reason we can rule out the possibility that the product is direct, i.e. we can assume that $C_P(H)$ is a proper subgroup of $P$.
Assume that the action is not faithful. Then $C_P(H)$ is a non-trivial normal subgroup of $Q_8$. This means that $K=P/C_P(H)$ is either cyclic of order two or the Klein four. So every non-trivial element of $K$ is of order two, and any two such elements commute. 
The action of every element of $K$, when viewed as a $2\times2$ matrix over $\Bbb{Z}_q$, is diagonalizable. Because they commute, they are simultaneously diagonalizable. Therefore, with respect to a carefully chosen basis, $K$ consists of diagonal matrices. Therefore the action of $K$, hence of $P$, on $H$ is not irreducible.

The above needed a bit of linear algebra: i) over a field of characteristic $\neq2$ a matrix of order two is diagonalizable, and ii) a commuting set of diagnolizable matrices is simultaneously diagonalizable. Recapping these bits to the extent required here.
If all the elements of $K$ act on $H$ as scalars, then any subgroup of $H$ is stable under the action of $P$ violating irreducibility. So we can assume that there exists an element $z\in K$ that does not act as a scalar on $H$. Necessarily $z$ has order two. 
For every element of $h\in H$ we have
$$h=\frac12(h+z\cdot h)+\frac12(h-z\cdot h).\qquad(*)$$
Let $$H^+=\{\dfrac12(h+z\cdot h)\mid h\in H\}\quad\text{and}\quad
H^-=\{\dfrac12(h-z\cdot h)\mid h\in H\}.$$
Clearly $H^+$ and $H^-$ are subgroups of $H$. By $(*)$ we have $H=H^+\oplus H^-$. Furthermore, $z\cdot u=u$ (resp. $z\cdot u=-u$) for all $u\in H^+$ (resp. for all $u\in H^-$). So $z$ acts as scalar $+1$ on $H^+$ and as $-1$ on $H^-$. By our assumption both $H^+$ and $H^-$ must be non-trivial. 
For any $k\in K$ we have $kz=zk$. If $u\in H^+$, then
$$
z\cdot(k\cdot u)=k\cdot(z\cdot u)=k\cdot u
$$
implying that $k\cdot u\in H^+$. Therefore $H^+$ is stable under the action of $K$ violating irreducibility.
