I'm reading Siegfried Bosch's Algebra From the Viewpoint of Galois Theory, which gives the following definition:
A finite field extension $L/K$ is said to be solvable by radicals if $L$ admits an extension field $E$ together with a chain of field extensions : $$ K =: E_0 \subset E_1 \subset \cdots \subset E_m $$ such that in each case, $E_{i+1}$ is obtained from $E_i$ by adjoining an element of the following type:
(1) a root of unity, or
(2) a zero of a polynomial of type $X^n − a \in E_i[X]$, where $\mathrm{char}(K) \nmid n$, or
(3) a zero of a polynomial of type $X^p − X − a \in E_i[X]$ for $p = \mathrm{char}(K) > 0$.
Then $L/K$ is necessarily separable.
Now I'm reading the proof of the following lemma:
Given a chain of finite field extensions $K \subset L \subset M$, the extension $M/K$ is solvable (resp. solvable by radicals) if and only if $M/L$ and $L/K$ are solvable (resp. solvable by radicals).
I am confused about the proof of the part "If $M/L$ and $L/K$ root equations are solvable by radicals, then $M/K$ root equations are solvable by radicals". The idea of this part of the proof is as follows:
Choose an extension $L'/L$ such that $L'/K$ can be exhausted by a chain of field extensions as specified in the above definition. Then consider the composite field $L'M$ in an algebraic closure of $M$ and use the fact that $L'M/L'$ is solvable by radicals. It follows that $L'M/K$ is solvable by radicals and that the same is true for $M/K$.
Where I get confused is why we get $L'M/K$ root solvable by radicals? I can't think of any reason.