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We have the chain of intermediate fields $$F=K_0\leq K_1 \leq \ldots \leq K_r=K$$

Let $\omega$ be the $n$th primitive root of $1$.

We consider the chain $$F\leq F(\omega)\leq F_1\leq \ldots \leq F_r=K(\omega)$$ with $F_i=K_i(\omega)$.

We have that $F_i=F_{i-1}K_i$ and $\text{Gal}(F_i/F_{i-1})\hookrightarrow \text{Gal}(K_i/K_{i-1})$.

Why does it hold that the exponent of $\text{Gal}(F_i/F_{i-1})$ divides the exponent of $\text{Gal}(K_i/K_{i-1})$ ?

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This has nothing to do with the specific structure of these groups - by the definition of "exponent" (for example, as the least common multiple of the orders of every element in the group), if $ H \subset G $ are finite groups, then the exponent of $ H $ always divides the exponent of $ G $...

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    Why does it hold that $\text{Gal}(F_i/F_{i-1})\subset \text{Gal}(K_i/K_{i-1})$ ? – 2017-01-21
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    It doesn't, but the former group embeds into the latter, so it is isomorphic to a subgroup of the latter; and the exponent of a group is preserved under isomorphism... – 2017-01-21
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    I got it!! Thank you very much!! :-) – 2017-01-21