Why normal subgroup chains in Galois theory I have began understanding Galois theory and I had a question regarding the relationships of normal subgroups to field extensions.
So given an irreducible polynomial over the rationals
$$a_1 + a_2x +\cdots+ a_nx^{n-1} + x^n $$
the initial group of symmetries of the roots $r_1,\ldots,r_n$ is the set of permutations of roots that uphold 
$$ \begin{pmatrix}
r_1 + r_2 +\cdots+ r_n = -a_n \\
r_1r_2 +r_1r_3 +\cdots+ r_{n-1}r_n  = a_{n-1} \\
\vdots \\
r_1r_2\cdots r_n = (-1)^n a_1
\end{pmatrix} $$ 
which of course is the symmetric group $S_n$.
Now if we take the field $\mathbb{Q}$ and extend it by one of the roots 
$$ \Bbb{Q} \rightarrow \Bbb{Q}(r_1)$$ 
The equations above obviously reduce to a smaller collection
$$ \begin{pmatrix}
r_2 + \cdots+ r_n = -a_n -r_1 \\
r_2r_3 + \cdots + r_{n-1}r_n  = -a_{n-1} - r_1{a_n}+r_1^2\\
\vdots \\
r_2 r_3 \cdots r_n = (-1)^n \frac{a_1}{r_1}
\end{pmatrix} $$
Which is preserved by the group $S_{n-1}$ of permutations of the roots.
So this sequence of permutation groups converges to the identity once enough field extensions have occurred.
$$ G \rightarrow G' \rightarrow G'' \rightarrow\cdots \rightarrow e $$ 
$$ \Bbb{Q} \rightarrow \Bbb{Q}(r_1) \rightarrow \Bbb{Q}(r_1)(r_2) \rightarrow\cdots\rightarrow \Bbb{Q}(\text{all roots}) $$ 
And I understand that
$$G' \subset G, \qquad G'' \subset G',\qquad\ldots $$ In other words each of those groups is a subgroup of its predecessor. But, it is not immediately clear to me why they need to be normal subgroups. 
Could someone explain that?

What you're asking about is false in general.
Let's set $F_0=\mathbb{Q}$ and $F_k=\mathbb{Q}(r_1,\ldots,r_k)$, so that $F_n=\mathbb{Q}(\text{all roots})$.
Now, $F_n/L$ is Galois for any field $\mathbb{Q}\subseteq L\subseteq F_n$ because $F_n$ is a splitting field. In particular, $F_n/F_k$ is a Galois extension for any $0\leq k\leq n$. Let's use proper notation to keep everything clear, and set $G_k=\mathrm{Gal}(F_n/F_k)$. Then we do have
$$G_0\supseteq G_1\supseteq\cdots\supseteq G_n=\text{trivial group}$$
but there is no reason why we would have $G_k\trianglerighteq G_m$ for any $k<m$ except for $k=0$ and $m=n$ (or rather, $m=n$ and any $k$ for which $F_k=F_n$). Certainly, the only case when we'd have $G_0\trianglerighteq G_m$ — that is, the only case when we'd have $\mathrm{Gal}(F_n/\mathbb{Q})\trianglerighteq\mathrm{Gal}(F_n/F_m)$ — is when $F_m/\mathbb{Q}$ is a Galois extension, which can't happen unless $F_m=F_n$, i.e., unless $F_m$ is the splitting field of the irreducible polynomial we started with.
For example, consider the polynomial $x^3-2$ with roots
$$r_1=\sqrt[3]{2},\qquad r_2=\zeta_3\sqrt[3]{2},\qquad r_3=\zeta_3^2\sqrt[3]{2}$$
Then we have $G_0=\mathrm{Gal}(\mathbb{Q}(\zeta_3,\sqrt[3]{2})/\mathbb{Q})\cong S_3$, but $G_1=\mathrm{Gal}(\mathbb{Q}(\zeta_3,\sqrt[3]{2})/\mathbb{Q}(\sqrt[3]{2}))$ is a subgroup of order $2$ in $S_3$, which will not be a normal subgroup.
