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?
 A: 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.
A: The link between (normal) subgroups and field subextensions of a Galois extension $L$, with Galois group $G$  is there is a bijection between subgroups of  $G$ and subextensions of $L$, defined by $\;H\mapsto L^H$ (the fixed points of $L$ under the action of $H$). 
Furthermore $L^H$ is Galois if and only if $H$ is a normal subgroup of $G$; in which case the Galois group of $L^H$ over $\mathbf Q$ is the quotient group $G/H$.
