I'm trying to show in an easy manner that if $K \subseteq M \subseteq L$, then the normal closure $N'$ of $L:M$ is contained in the normal closure $N$ of $L:K$.
My book uses the following definition of normal closure: $N$ is a normal closure of $L:K$ if $N$ is a finite extension of $K$ containing $L$ such that:
$1)$ $N:K$ is normal, and
$2)$ If $L \subseteq M \subseteq N$ and $M:K$ is normal then $M=N$.
The proof I know is the following:
Let $A=${$a_1, ..., a_r$} be a basis for $L:M$ such that $1=a_1 \in A$, and
let $B=${$b_1, ..., b_s$} be a basis for $M:K$ such that $1=b_1 \in B$, then
$AB = $ {$ab:a \in A, b \in B$} $=C =${$c_1, ..., c_n$} is a basis for $L:K$ such that $A \cup B \subseteq C$.
Let $p_i$ be the minimal polynomial of $a_i$ over $M$ and let $f_j$ be the minimal polynomial of $c_j$ over $K$, then $N'$ is the splitting field of $ \prod ^n _{i=1} p_i$ over $M$, while $N$ is the splitting field of $ \prod ^n _{j=1} f_j$ over $K$.
Let $x$ be any zero of $p_i$. Such $p_i$ is the minimal polynomial of $a_i$ over $M$, notice that $a_i \in C$, consider then the minimal polynomial $f_j$ of $c_j = a_i$ over $K$. In $M[t]$ we have $p_iq=f_j$, so that all the zeroes of $p_i$ are zeroes of $f_j$. But all the zeroes of $f_j$ are contained in $N$ by definition. Since the zeroes of the $f_i$ generate $N'$, we must have $N' \subseteq N$.
The above proof is nowhere in my book. The issue as to why I'm seeking a different, easier proof is that, while proving a certain theorem in the book, the author implicitly assumes $N' \subseteq N$, leading me to wonder if there is an intuitive, shorter explanation of said fact.
I would really appreciate any help.