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 $\text{Gal}(K_i/K_{i-1})$ ?


1 Answer 1


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 $...

  • $\begingroup$ Why does it hold that $\text{Gal}(F_i/F_{i-1})\subset \text{Gal}(K_i/K_{i-1})$ ? $\endgroup$
    – Mary Star
    Jan 21, 2017 at 22:36
  • 1
    $\begingroup$ 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... $\endgroup$
    – Ege Erdil
    Jan 21, 2017 at 22:40
  • $\begingroup$ I got it!! Thank you very much!! :-) $\endgroup$
    – Mary Star
    Jan 21, 2017 at 23:20

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .