Let$G$ be a group of order $12$ with $4$ Sylow $3$-subgroups, affording permutation representation $\phi:G \to S_4$. Prove that $ker \phi = 1$
This is what I am trying to understand from Dummit and Foote's Abstract Algebra 3rd edition page 144 chapter 4.4. The explanation in the book goes:
Since distinct Stlow 3-subgrouos intersect in the identity and each contains two elements of order 3, $G$ contains $2•4=8$ elements of order $3$. Since $|G:N_G(P)| = n_3 = 4, N_G(P) = P$. Now $G$ acts by conjugation in its four Sylow 3-subgroups, so this action affords a permutation representation $\phi : G \to S_4$. The kernel $K$ of this action is the subgroup of $G$ which normalizes all Sylow $3$-subgroups of $G$. In particular, $K \leq N_G(P) = P$. [[[[[Since $P$ is not normal in $G$ by assumption, $K=1$]]]]
I understand everything but the bracketed portion. I do not understand wht $K$ cannot be $P$, though I do inderstand that $K$ has to be either $P$ or $1$.
Firstly, note that the abelian groups of order 12 are $\mathbb{Z}_6\times \mathbb{Z}_2$ and $\mathbb{Z}_{12}$. The first one has only 2 elements of order 3, so there is only one subgroup of order 3. The second one has only one subgroup for each divisor of 12 since is cyclic (see this property in reference (*)). Thus our group is not abelian, that is to say, is isomorphic to the dihedral group $D_6$, the alternating group $A_4$ or the dicyclic group $\text{Dic}_3$.
Now, we consider the action of $G$ given by $$\begin{equation*} \begin{split} \phi : G & \to \text{Bijection}(X)=S_4 \\ g &\mapsto \overline{g} \end{split} \end{equation*}$$ where $X=G/R_H$ and $R_H$ is the equivalence relation which induces the right coset on a subgroup H, and $$\begin{equation*} \begin{split} \overline{g}: X & \to X \\ Hx &\mapsto Hxg \end{split} \end{equation*}$$ It's easy to check that $\overline{g}$ is bijective and its inverse is $$\begin{equation*} \begin{split} \overline{g}^{-1}: X & \to X \\ Hy &\mapsto Hxg^{-1} \end{split} \end{equation*}$$ Also, it's easy to check it's in fact an action. Now, we observe that $\text{ker}\phi \subset H$ because each element $u\in \text{ker}\phi$ verifies that $$H=\overline{u}(H)=Hu,$$ so $u\in H$. In particular, $\text{ker}\phi \subsetneq G$ and, as $\text{ker}\phi \vartriangleleft G$, we deduce that if $\phi$ is not injective the group $G$ is not simple.
In our case, as $\text{ord}(H)=3$, we have two options: $\text{ker}\phi=\text{id}$ or $\text{ker}\phi=H$ (because groups of prime order haven't got any proper subgroup)
If $\text{ker}\phi=H$, then $H$ is the unique $3$-Sylow subgroup (because it's normal), and that contradicts the fact that $G$ has $4$ $3$-Sylow subgroups. Then, $\text{ker}\phi=\text{id}$. This case corresponds with the alternating group $A_4$. In fact, this is unique subgroup of order 12 of $S_4$.
(*) In a finite cyclic group there is only one subgroup of order $k$.