I study a lemma from 'Local Class Field Theory' and I have difficulties.

To understand this properly we need another lemma


Let $\mathfrak{l=o/p}=\mathbb{F}_q$ for a $p$-field $(k,v)$. Then, for each $x \in \mathfrak{o}$, the limit $\omega(x) = \lim_{n\to\infty}x^{q^n}$ exists in $\mathfrak{o}$, and the map $\omega :\mathfrak{o} \to \mathfrak{o} $ has the following properties: $\omega(x)\equiv x \ mod \ \mathfrak{p}$, $ \ $ $\omega(x)^q = \omega(x) \ $, $\omega(xy)=\omega(x)\omega(y)$.

$\mathfrak{o}=\{x \in k | v(x) \geq 0 \}$, $\mathfrak{p} =\{x \in k | v(x) > 0\}$.

Lemma 2:

Let $(k,v)$ be as stated above and let $$ V=\{ x\in k | x^{q-1} =1 \}, \ A=V \cup \{0\} = \{x \in k| x^q=x\}.$$ Then $A$ is a complete set of representatives of $\mathfrak{l=o/p} $ in $\mathfrak{o} $,containing $0$; $V$ is the set of all $(q-1)$st roots of unity in $k$; and the canonical ring homomorphism $\mathfrak{o \to l=o/p}$ induces an isomorphism of multiplicative groups: $$ V \cong l^{\times}$$ In particular, $V$ is a cyclic group of order $q-1$.

Proof of Lemma 2:

Let $A' = \{\omega(x) | x\in \mathfrak{0} \}$. As $\omega(x) \equiv x \ mod \ \mathfrak{p}$, each reside class of $\mathfrak{o}$ mod $\mathfrak{p}$ contains at least one element in $A'$, and as $\omega(x)^q =\omega(x)$, $A'$ is a subste of $A$. However, the number of elements $x$ in $k$ satisfying $x^q-x=0$ is at most $q$, while the number of elemenst in $\mathfrak{l=o/p}$ is $q$. Hence $A=A'$ and $A$ is a complete set of representatives of $\mathfrak{l=o/p}$ in $\mathfrak{o}$. Obviously $0=\omega(x) \in A$. Since $\omega(xy)=\omega(x)\omega(y)$, the other statements on $V$ are clear. QED

My questions:

1) What does it mean that homomorphism induces an isomoprhism?

2) Why we need to consider equation: $x^q-x=0$?

3) Why $\omega(xy)=\omega(x)\omega(y)$ implies the other statements ? (I understand it implies that map $ \omega: \mathfrak{o} \to \mathfrak{o}$ is homomorphism)


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