The general statement of what you're heading toward is...
$G$ has a normal subgroup of prime index $p$ if and only if there exists a non-zero homomorphism from $G$ to $\mathbb{Z}_p$.
Proof: Suppose $H$ is a normal subgroup of index $p$. Then $G/H$ is a subgroup of order $p$. Thus it's isomorphic to $\mathbb{Z}_p$ since $p$ is prime: call this isomorphism $\varphi$. Let $\pi:G \to G/H$ be the projection homomorphism $\pi(g)=gH$. Then $\varphi \circ \pi$ is a non-zero homomorphism from $G$ to $\mathbb{Z}_p$.
Now suppose there is a non-zero homomorphism. Since any non-zero element of $\mathbb{Z}_p$ generates the group, this homomorphism is onto and so by the isomorphsm theorem $G/\mathrm{Ker}$ is isomorphic to $\mathbb{Z}_p$ which implies that the kernel (a normal subgroup) has index $p$.
Now if there is a surjective homomorphism onto $\mathbb{Z}_n$ then one has a normal subgroup of index $n$ (by the isomorphism theorem). However, if one has a normal subgroup of index $n$ (not nec. prime), then there is a surjective homomorphism onto a group of order $n$, but in general this group doesn't have to be cyclic (or even abelian).
Now if you want to probe subgroups of index $n$ which are not necessarily normal, your line of inquiry probably won't pan out -- because you're essentially looking at kernels (which are always normal). To look at subgroups (not nec. normal), group actions are the way to go.
We say $G$ acts on a set $S$ if there is a map $\cdot:G \times S \to S$ denoted $(g,s) \mapsto g \cdot s$ such that $(gh)\cdot s = g \cdot (h \cdot s)$ for all $g,h \in G$ and $s \in S$ and also $e \cdot s= s$ for all $s \in S$ if $e$ is the identity of $G$.
Let $s \in S$. Then $G \cdot s = \{ g \cdot x \;|\; g \in G\}$ is the orbit of $s$ and $G_s = \{ g \in G \;|\; g \cdot s = s \}$ is the isotropy subgroup fixing $s$ (or stabilizer of $s$). It's not hard to show that isotropy subgroups are in fact subgroups and the orbits of elements partition the set being acted upon. It can be shown that $[G:G_s]=|G \cdot s|$ (sort of a generalization of Lagrange's theorem), so the size of the orbit of $s$ times the size of the stabilizers of $s$ equals the size of the group.
Now suppose $H$ is a subgroup of $G$ of index $n$, then $G$ acts transitively on the left cosets of $H$ in $G$ (which is a set with $n$ elements): $g \cdot xH = (gx)H$. [Transitive means that there is only 1 orbit.]
Conversely, if $G$ acts transitively on a set of size $n$ then any isotropy subgroup (stabilizer) is a subgroup of index $n$ since the index of an isotropy subgroup fixing $x$ is equal to the size of the orbit of $x$ which is everything (i.e. $n$ elements) if we assume the action is transitive.
The second to last line is where you reach a problem: We don't have an "image" of $S_3$, we have a pre-image. This pre-image contains $P_3$, but is not in necessity equal to it. In fact, since the Slyow 3-group of $S_3$ contains the identity, the pre-image of the subgroup of $S_3$ contains the kernel, properly. But the pre-image of a normal subgroup is normal. So we must only really show that the pre-image is a proper subgroup.
A simpler proof though, notes that we may let $\phi$ be the representation by conjugation on the Slyow 2-groups (which are of order 16). If this is trivial, then we see that $n_1=1$ by Slyow's Theoroms. If not, then Ker $\phi$ is the desired proper normal subgroup (Since Ker $\phi=0$ makes no sense considering the order).
Best Answer
Consider the chain of homomorphisms $$ G\xrightarrow{\pi}G/N\xrightarrow{\varphi}\mathbb{Z} \xrightarrow{p_n}\mathbb{Z}/n\mathbb{Z} $$ where $\varphi_n=p_n\circ\varphi\circ\pi$ is surjective. Then consider the induced homomorphism $$ G/\ker\varphi_n\to\mathbb{Z}/n\mathbb{Z} $$ What does it mean that a normal subgroup $K$ of $G$ has index $n$? That the quotient group $G/K$ has $n$ elements.
So, what can we say about $\ker\varphi_n$? The answer is in the homomorphism theorem: the induced homomorphism $G/\ker\varphi_n\to\mathbb{Z}/n\mathbb{Z}$ is injective by construction and also surjective because $\varphi_n$ is surjective. So it is an isomorphism and, since $|\mathbb{Z}/n\mathbb{Z}|=n$, $\ker\varphi_n$ is a normal subgroup of $G$ of index $n$ (and containing $N$).