Exercise 1.7.7 (The action $(\alpha, v)\mapsto \alpha v$ of Example 2 is faithful )

Question

Prove that in Example 2 in this section the action is faithful.

Richard Ganaye · Accepted Answer

\begin{proof} 
The action of example 2 is the map
$$
\left\{
\begin{array}{ccl}
F^* 	imes V & 	o & V \
(\alpha, v)& \mapsto &\alpha \cdot v = \alpha v
\end{array}
ight.
$$
where $V$ is a $F$-vector space. (We suppose that $V 
e\{0\}$ is not the trivial vector space, otherwise the action is not faithful.)

The corresponding permutation representation is
$$
\varphi
\left\{
\begin{array}{ccl}
F^* & 	o &S_V\
\alpha & \mapsto &\varphi_\alpha,
\end{array}
ight.
\qquad 	ext{where} \qquad \varphi_\alpha
\left\{
\begin{array}{ccl}
V & 	o &V\
v & \mapsto &\varphi_\alpha(v) = \alpha v.
\end{array}
ight.
$$
Then 
\begin{align*}
\alpha \in \ker(\varphi) & \iff \varphi_\alpha = \mathrm{Id}_V\
&\iff \forall v \in V,\ \alpha v = v & 	ext{(1)}.
\end{align*}
Since $V 
e \{0\}$, there is some vector $v_0 \in V,\ v_0 
e 0$. If $\alpha \in \ker(\varphi)$, then by (1), $\alpha v_0 = v_0$.

Assume for the sake of contradiction that $\alpha 
e 1$. Then $(\alpha - 1) v_0 = 0$, thus $(\alpha-1)^{-1} (\alpha-1) v_0 = 0$, thus $v_0 = 0$. This is contradiction, which proves that $\alpha = 1$. Therefore $\ker(\varphi) = \{1\}$, so the action is faithful.
\end{proof}

Exercise 1.7.7 (The action $(\alpha, v)\mapsto \alpha v$ of Example 2 is faithful )

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Exercise 1.7.7 (The action $(\alpha, v)\mapsto \alpha v$ of Example 2 is faithful )

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