Definition 2.15. A group action of $G$ on $X$ is called faithful (or effective) if different elements of $G$ act on $X$ in different ways: when $g_1\neq g_2$ in G, there is an $x\in X$ such that $g_1\cdot x \neq g_2\cdot x$.
Example 2.17. The action of $G$ on itself by conjugation is faithful if and only if $G$ has a trivial center, because $g_1gg_1^{-1} = g_2gg_2^{-1}$ for all $g\in G$ if and only if $g_2^{-1}g_1$ is in the center of $G$. When $D_4$ acts on itself by conjugation, the action is not faithful since $r^2$ acts trivially (it is in the center), so $1$ and $r^2$ act in the same way.
Source: Keith Conrad, Group Actions.
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I don't understand how you calculate $g_2^{-1}g_1 \in Z(G)$. It looks like the author is working with the negation of the definition of faithful? Then we need to prove: $$g_1 \cdot x = g_2 \cdot x \implies g_1 = g_2.$$
I understand that because $r^2 = \text{two $90^\circ$ rotations}$, hence $r^2g(r^2)^{-1} = g$ for all $g \in G$.
The author also presupposes $g = x \in G$. Hence
$$
\begin{align}
g_1 \cdot x = g_2 \cdot x & \iff g_1gg_1^{-1} = g_2gg_2^{-1} \\
& \iff \color{magenta}{g_2^{-1}}g_1gg_1^{-1} = \color{magenta} {g_2^{-1}}g_2gg_2^{-1} \\
& \iff \color{magenta}{g_2^{-1}}g_1gg_1^{-1} = gg_2^{-1}.
\end{align}
$$What now?
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How do you realize one have to left-multiply by $\color{magenta}{g_2^{-1}}$?
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What's the intuition?
Best Answer
Regarding (1.) I understand that left multiplying by $g_2^{-1}$ and right multiplying by $g_1$ gives $$ g_1 g g_1^{-1} = g_2 g g_2^{-1} \quad \forall g \qquad \rightarrow \qquad (g_2^{-1} g_1) g = g (g_2^{-1} g_1) \quad \forall g $$ wich is the definition of $g_2^{-1} g_1$ being in the center of $G$. Then only if the center is trivial, i.e. $g_2^{-1} g_1= e$ or equivalently $g_1 = g_2$, then the action is faithful.