The holomorph of $Z_2 \times Z_2$ I want to solve the following problem from Dummit & Foote's Abstract Algebra text (p. 184, Exercise 5):

Let $G=\text{Hol}(Z_2 \times Z_2)$
(a) Prove that $G=H \rtimes K$ where $H=Z_2 \times Z_2$ and $K \cong S_3$. Deduce that $|G|=24$.
(b) Prove that $G$ is isomorphic to $S_4$. [Obtain a homomorphism from $G$ into $S_4$ by letting $G$ act on the left cosets of $K$. Use Exercise 1 to show that this representation is faithful.]

Exercise 1 basically states that if $G=H \rtimes_\varphi K$ then $C_K(H)= \ker \varphi$.
My attempt:
(a) By definition $\text{Hol}(Z_2 \times Z_2)=(Z_2 \times Z_2) \rtimes_{\text{id}} \text{Aut}(Z_2 \times Z_2)$. Since the automorphism group of the Klein 4-group is isomorphic to $S_3$ (a previous result) we are done.
(b) Let $G$ act on $G/K$ by left multiplication, and let $\pi_K:G \to S_{G/K} \cong S_4$ be the induced permutation representation. We have that $$\ker \pi_K=\bigcap_{g \in G} gKg^{-1}$$ is the normal core of $K$... 
This is where I'm stuck, since the homomorphism $\varphi$ in the semidirect product $G$ is the identity, Exercise 1 gives $C_K(H)=1$. I was thinking of showing that the action is faithful by proving $\ker \pi_K \leq C_K(H)(=1)$, but I couldn't do it.
Can anyone please help me solve part (b) using Exercise 1 as hinted by the authors?
Thank you!
 A: Here is my attempt at a solution. Glaring, painfully obvious errors are possible.
Suppose $g \in$ ker $\pi_H$. Then we would have $gxK=xK$ for every $x \in G$. In particular, note that we must have $g \in K$ (setting $x = 1$). 
$$gxK=xK \enspace\forall x\in G\Leftrightarrow \\x^{-1}gx \in K \enspace \forall x\in G \Leftrightarrow \\x^{-1}gxg^{-1} \in K \enspace \forall x\in G$$
Where the final line is because $g$ (and therefore $g^{-1}$) $\in K$. Restricting $x$ to only $H$ instead of all of $G$, we must have that $$x^{-1}gxg^{-1} \in K \enspace \forall x \in H$$ By normality of $H$, we must also have that $x^{-1}gxg^{-1} \in H \enspace \forall x \in H$, but since $H \cap K = 1$, $x^{-1}gxg^{-1} =1 \enspace \forall x \in H$, or $gx=xg \enspace \forall x \in H$. This means that $g \in C_K(H)$, but since $C_K(H)=1$, $g$ was the identity this whole time.
A: Ok, perhaps now I can be of some help: since in the holomorph the homomorphism is the identity one, we have that $\;\ker\varphi=1\;$ , from which 
$$\;C_{\text{Aut}(\Bbb Z_2\times\Bbb Z_2)}(\Bbb Z_2\times\Bbb Z_2)=1\iff \text{the only automorphism centralizing}\;\;\Bbb Z_2\times\Bbb Z_2\;$$
is the identity one (all this happens in the holomorph, of course, but it follows at once from the fact that the only permutation leaving fixed all the elements is the identity one) .
Thus, in the action of $\;G\;$ on the left cosets of $\;K\;$ (and thus in the induced homomorphism $\;G\to S_4\;$), we get that $\;x\cdot(gK):=(xg)K=gK\iff g^{-1}xg\in K\;$ and the last expression is true as inner automorphisms are automorphisms (this is painfully tautological).
Thus, the homomorphism is a monomorphism (i.e., the action is faithful) and we're done.
