I am now studying Galois theory from NPTEL online lecture course on Galois theory. While studying the lecture notes I found some statements in the proof of the theorem stated in the title which I failed to understand. Here's the proof given by Prof. Dilip P. Patil of IIT Bombay https://youtu.be/MMTha7gkr_o?list=PLOzRYVm0a65dsCb_gMYe3R-ZGs53jjw02&t=1219
Let me copy the proof what he had given.
Theorem $:$ Every finite subgroup $G \subseteq \Bbb K^{\times},$ $\Bbb K$ field is cyclic.
Proof $:$ Let $m=|G|={p_1}^{\alpha_1} {p_2}^{\alpha_2} \cdots {p_r}^{\alpha_r} = m_1m_2 \cdots m_r,$ where $m_i = {p_i}^{\alpha_i}$ for $i=1,2,\cdots, r$ where $p_1,p_2,\cdots,p_r$ are distinct primes and $\alpha_1,\alpha_2,\cdots,\alpha_r \in \Bbb N.$ So $m_1,m_2,\cdots,m_r$ are pairwise relatively prime to each other.
Let $n_i=\frac {m} {m_i} \in \Bbb N,$ for $i=1,2,\cdots,r.$ Then $\text{gcd} (n_1,n_2,\cdots,n_r)=1.$
Write $1=b_1n_1+b_2n_2+\cdots+b_rn_r,\ b_1,b_2,\cdots ,b_r \in \Bbb Z.$ Then $\forall x \in G$
\begin{align*} x = x^1 &= x^{b_1n_1+b_2n_2+\cdots+b_rn_r}\\ & = \left (x^{b_1n_1} \right )\left (x^{b_2n_2} \right ) \cdots \left (x^{b_rn_r} \right )\\ &= x_1x_2\cdots x_r. \end{align*}
where $x_i=x^{b_in_i},$ for $i=1,2,\cdots, r.$
Now for each $i=1,2, \cdots , r$ we have ${x_i}^{m_i} = \left (x^{b_in_i} \right )^{m_i} = x^{b_im} = (x^m)^{b_i} = 1,$ by Lagrange's theorem. i.e. $x_1,x_2,\cdots,x_r$ are zeros of the polynomial $X^{m_i}-1.$ i.e. $X^{m_i} – 1$ has exactly $m_i$ zeros in $G$ (for otherwise $G$ has less than $m=m_1m_2\cdots m_r$ many elements) for each fixed $i=1,2,\cdots,r.$
But $X^{\frac {m_i} {p_i}} – 1$ has at most $\frac {m_i} {p_i} < m_i$ zeros in $G$ for $i=1,2,\cdots ,r.$ So $\exists$ $y_i \in G$ such that ${y_i}^{m_i} = 1$ but ${y_i}^{\frac {m_i} {p_i}} \neq 1$ for $i=1,2,\cdots , r.$
i.e. $\exists$ $y_i \in G$ such that $\text {Ord}\ (y_i) = m_i$ for $i=1,2,\cdots ,r.$
So, \begin{align*} \text{Ord}\ (y_1y_2\cdots y_r) &= \text {Ord}\ (y_1) \text {Ord}\ (y_2) \cdots \text {Ord}\ (y_r)\\ &=m_1m_2\cdots m_r = m = |G|.\end{align*}
So, $y_1y_2\cdots y_r$ is a generator of $G.$
Therefore, $G$ is a cyclic group.
QED
But I failed to understand why each $x_i$ is a root of the polynomial $X^{m_i}-1.$ It is only proved that $x_i$ is a zero of $X^{m_i} – 1$ for $i=1,2, \cdots , r.$ But for $j \neq i$ how can $x_j$ be a root of $X^{m_i} – 1$ for $i=1,2, \cdots , r$? Also I failed to understand the argument that $X^{m_i}-1$ has exactly $m_i$ zeros in $G$ for $i=1,2,\cdots , r.$
Any help in understanding those arguments made in the above proof will be highly appreciated. Thank you very much.
Best Answer
You don't need that any $x_j$ be a solution of $X^{m_i}-1$: the point is that any $x\in G$ can be decomposed as a product $\prod_{i=1}^r x_i$, where each $x_i$ solves $X^{m_i}-1$ (for the same $i$). Hence, if $k_i$ is the number of elements in $G$ solving $X^{m_i}-1$, $G$ has at most $\prod_{i=1}^r k_i$ elements, and so any $k_i$ must equal $m_i$: from the fact that $X^{m_i}-1$ has at most $m_i$ solutions in $\mathbb{K}$ you get $k_i\leq m_i$, but you need $\prod_{i=1}^r k_i\geq\prod_{i=1}^r m_i$, and therefore $k_i=m_i$ for all $i$.