Classifying the representations of $G=\langle(123), (456), (23)(56) \rangle \subset S_6$. This is exercise 3 of chapter 2 of the book "Representations of finite and lie groups" By B. Thomas. My questions are in bold. I am essentially confused by two steps but I think I understand the entire question apart from that.

Let G be a subgroup of order $18$ in the symmetric group $S_6$ given by
$$ G = \langle(123), (456), (23)(56)\rangle.$$
Show that G has a normal subgroup of order $9$ and four normal subgroups
of order $3$. By considering quotients show that G has two representations of degree one and four inequivalent irreducible representations of degree $2$, none of which is injective/faithful.


My approach is as follows.
Observe  $|G|=18=2 \cdot 3^2$.  Denote $n_3$ for the number of Sylow $3$-subgroups. We see that by Sylow's theorem $n_3 \equiv 1 \bmod 3$, so candidates are $n_3=1, 4, 7$  etc. However we also have that $n_3|2$, so $n_3=1.$  This means there is one sylow subgroup of order $9$, and another consequence of the Sylow theorems is that this group is then normal. We can construct such a group by considering the two generators of order $3$:
$$ H= \langle(123), (456) \rangle  \cong C_3 \times C_3.$$
Now this group has index $|G:H|= \frac{|G|}{|H|}=18/9=2$ and therefore it must indeed be normal in $G$. We can also use the generators (and one of its inverses) of order $3$ to construct $4$ cyclic subgroups of order $3$.
$$ N_1=\langle (1 2 3)\rangle=\{e, (123), (132)\} \cong C_3$$
$$ N_2=\langle (4 5 6)\rangle=\{e, (456), (465)\}  \cong C_3$$
$$ N_3=\langle (1 2 3)(4 5 6)\rangle=\{e, (1 2 3)(4 5 6), (1 3 2)(4 6 5)\} \cong C_3$$
$$ N_4=\langle (1 2 3)(4 6 5)\rangle=\{e, (1 2 3)(4 6 5), (1 3 2)(4 5 6 )\}  \cong C_3$$
We have thus found four subgroups of order $3$, but we still have to prove they are normal in $G$.

(how do I know these exhaust all the subgroups of $G$ one can have of order $3$? I do not know how to formulate this/the exact reason). I also thought earlier I knew why these were normal, but I was mistaken.

We can now use the fact that we may lift representations of quotients to the original group using the canonical quotient homomorphism.
$|G/H|=2$ and it is an abelian group (specifcally $C_2$) therefore by a corollary/consequence of Weddderburn's theorem we must have $2$ irreducible complex representations of $G/H$, which are both $1$ dimensional (I prefer to use Dummit and Foote corollary 11(1), page 861 for this). We lift these to $G$.
Now we can consider the quotients $$G/N_1, G/N_2, G/N_3, G/N_4 $$

According to the hints and solutions section in the back, these each give a $2$-dimensionial representation, why?

Once we accept this we can also lift these $4$ representations to $G$ to end up with $4$ (distinct/inequivalent) representations of $G$. The dimensions of the representations of $G$ we have so far are thus $1, 1, 2, 2, 2, 2$. Then observe we must also have by theorem $10.(4)$ of page 861 of Dummit and Foote that:
$$ \sum_{i=1}^r n_i ^2 = |G $$
However if we compare this to what we already have, we find:
$$ 2 \cdot 1^2 + 4 \cdot 2^2 =18 =|G|$$
So we indeed find that these are all representations by exhaustion. A lift of a quotient $G/N$ to $G$ always has $N$ contained in its kernel. We therefore know that lifted representations never have a trivial kernel and therefore they are not faithful (since they are not injective homomorphisms).
 A: To answer my own questions in bold:
The list of elements in $N_1$, $N_2$, $N_3$ and $N_4$ exhaust all elements of order $4$ in $G$ and therefore these are the only four groups we can form of order $4$.
We can compute normality directly by looking what conjugation by the generators does to each of these subgroups. One can compute directly for $g\in \{(123),(456), (23)(56)\}$ that:
$$ g N_i g^{-1}\subseteq N_i$$
For instance:
$$(456)N_1(456)^{-1}= (456)\{e, (123), (132)\}(465) =$$ $$\{(456)(465),(456)(123)(465), (456)(132)(465)  \}= \{e, (123), (132)\}=N_1 $$
And then for a product of two  generators $a, b$ we have:
$$(ab)N_i (ab)^{-1}= a(bN_ib^{-1})a^{-1}\subseteq aN_i a^{-1} \subseteq N_i. $$
So if we know that the subgroups are left invariant under conjugation by the generators, the conjugation action of the entire group on the subgroups $N_i$ leaves these subgroups invariant, therefore they are normal subgroups by the subgroup test.
If we look at the quotients $G/N_i $ we notice they are of order $18/3=6$. There are two groups (up to isomorphism) of order $6$, that is $S_3$ (non-abelian) and $C_6$(abelian, cyclic). In $G$ we get that:
$$(23)(56) (456)= (23) (4 6).$$
But also:
$$(456) (23)(56)=(456)(56)(23)= (45)(23). $$
In the quotient $G/N_1$ these elements do not get mapped to $0$ or to the same elements. We therefore know that the quotient is not abelian. One can also do a similar check for the other quotients or investigate the order of the elements and realise that no element has order $6$. Upon inspection all the quotients have the structure of $S_3$, which has three irreps of dimension $1, 1, 2$ as was one of the examples in the lecture notes/textbook.
