Proof of lemma regarding the number of roots of unity The following is from A. Białynicki-Birula, "Algebra" (the translation is mine). I put the part that confuses me in >>> ... <<<.

Theorem 7.6. Let $e$, $e'$ be elements of a given field $K$. Then:
  (c) if $e \in \mu_n(K)$, $e' \in \mu_m(K)$, then $e \cdot e' \in \mu_r(K)$, where $r = lcm(n,m)$,
  ...
  (e) if $e \in \mu_{n \cdot m} (K)$, then $e^n \in \mu_m(K)$.
Lemma 7.10. Let $K$ be a field containing exactly $n$ different $n$-th roots of unity, $m \in N_{+}$ and $m | n$. Then: the field $K$ contains exactly $m$ different $m$-th roots of unity.
Proof. Let $n = m \cdot m_1$ and let $card(\mu_m(K)) := m'$, $card(\mu_{m_1}(K)) := m'_1$. Let $\phi : \mu_n(K) \rightarrow \mu_m(K)$ be a mapping given by $\phi(e) := e^{m_1}$ (compare Theorem 7.6(e)). If $e, e' \in \mu_n(K)$ and $\phi(e) = \phi(e')$, then
$$(e'e^{-1})^{m_1}=(e')^{m_1} \cdot (e^{-1})^{m_1} = 1$$
thus $e'e^{-1} \in \mu_{m_1}(K)$. Conversely, if $e \in \mu_n(K)$ and $f \in \mu_{m_1}(K)$, $e' = fe$, then $e' \in \mu_n(K)$ and $\phi(e) = \phi(e')$. Thus >>> $card(\phi^{-1}(f)) = m'_1$ <<<. Let us break the set $\mu_n(K)$ into sets of the form $\phi^{-1}(f)$, $f \in \mu_m(K)$. As there will be at most $m'$ such sets (because $card(\mu_m(K)) = m'$) and each of these sets has $m'_1$ elements, so $n \leq m' \cdot m_1'$. However, since $m' \leq m$, $m'_1 \leq m_1$, we obtain $m' =m$ and $m'_1 = m_1$ which completes the proof.



*

*Does saying about $\phi^{-1}(f)$ actually make sense? The values of $\phi$ are by the very definition of this mapping in $\mu_m(K)$ while the author assumes $f$ to be in $\mu_{m_1}(K)$.

*If $\phi^{-1}(f)$ indeed makes sense, then how did he conclude that it has $m'_1$ elements? From what I can see he established a bijective mapping of some sort, but I got lost trying to figure out the details.

*If that proof is broken, is it possible to (fairly) easily rescue it?

 A: The expression $\phi^{-1}(f)$ certainly does not make sense in the context of the formulation 

if $e \in \mu_n(K)$ and $f \in \mu_{m_1}(K)$, $e' = fe$, then $e' \in \mu_n(K)$ and $\phi(e) = \phi(e')$. Thus $\text{card}(\phi^{-1}(f)) = m'_1$.

However, I believe that the author didn't mean the above $f \in \mu_{m_1}(K)$, but meant it in the sense of the next sentence

Let us break the set $\mu_n(K)$ into sets of the form $\phi^{-1}(f)$, $f \in \mu_m(K)$.

I would say that his notation is confusing and careless. But you will find many such examples in the literature. So let us be very precise and use better notation.
For a fixed $e \in \mu_n(K)$ the author establishes a bijection between $\phi^{-1}(\phi(e)) = \{ e' \in \mu_n(K) \mid \phi(e) = \phi(e') \}$ and $\mu_{m_1}(K)$. In fact, there is a well-defined function $u : \phi^{-1}(\phi(e)) \to \mu_{m_1}(K)$ given by $u(e') = e'e^{-1}$. Also the function $v : \mu_{m_1}(K) \to \phi^{-1}(\phi(e))$ given by $v(f_1) = f_1e$ is well-defined because $(f_1e)^n = f_1^n e^n = ((f_1)^{m_1})^m e^n = 1$ and $\phi(f_1e) = (f_1e)^{m_1} = f_1^{m_1}e^{m_1} = e^{m_1} =  \phi(e)$. But now $v(u(e')) = v(e'e^{-1}) = e'e^{-1}e = e'$ and $u(v(f_1)) =u(f_1e) = f_1ee^{-1} = f_1$ which shows that $u, v$ are inverse bijections.
This means that for any $f \in \phi(\mu_n(K))$ the set $\phi^{-1}(f)$ has $m'_1$ elements. Since at the moment we do not know whether $\phi$ is surjective, we can only say that $\phi(\mu_n(K))$ has $r \le m'$ elements. This implies $n = r \cdot m_1 \le m' \cdot m'_1$. I am sure that the author has proved somewhere that $\text{card}(\mu_k(K)) \le k$, so we have $m' \le m$ and $ m'_1 \le m_1$. If $m' < m$ or $ m'_1 < m_1$, then $m'm'_1 < mm_1 = n$ which contradicts $n \le m' \cdot m'_1$. Therefore $m' = m$ and $ m'_1 = m_1$.
This also shows that $\phi$ is surjective. If it were not, then $\phi(\mu_n(K))$ would have $r < m$ elements and $\mu_n(K)$ would have $r \cdot m_1 < m \cdot m_1 = n$ elements.
Remark: There is a variant of the proof which is in my opinion is more elegant. The sets $\mu_k(K)$ are subgroups of the (abelian) group $(K \setminus \{0\},\cdot)$. Perhaps the author has proved it somewhere, otherwise it can be done very easily. The map $\phi$ is a group homomorphism, thus it induces a group isomorphism $\phi' : \mu_n(K) / \ker(\phi) \to \text{im}(\phi) = \phi(\mu_n(K))$. We have $\ker(\phi) = \mu_{m_1}(K)$ because $\mu_{m_1}(K) \subset \mu_n(K)$ and $e' \in \ker(\phi)$ is equivalent to $(e')^{m_1} = 1$. Thus  $\mu_n(K) / \mu_{m_1}(K) \approx \text{im}(\phi)$. This implies
$$(*) \quad n/m'_1 = \text{card}(\mu_n(K) / \mu_{m_1}(K)) = \text{card}(\text{im}(\phi)) \le m' \le m = n/m_1 .$$
Hence $m'_1 \ge m_1$. We conclude $m'_1 = m_1$ because by definition $m'_1 \le m_1$. Inserting this in $(*)$ shows that $m' = m$ (and $\text{card}(\text{im}(\phi)) = m$ which means that $\phi$ is surjective).
