The group G acts faithfully on X

Question

Prove that $G$ acts faithfully on $X$ when there are no two elements of $G$ acting the same way on an element $X$.

So I don't have much of a proof, but here's what I'm thinking. I know that for $G$ to act faithfully on $X$, the identity is the only element that fixes every element in $X$, so $\forall x \in X, ex = x$. So this means that, $\forall g \in G, gx \neq x$. So if $\phi$ is the action, $ker\phi = \{e \in G | ex = x\}$. I don't know how to derive the proof of this though.

Answer 1

Let $g,h\in G$.
Suppose $gx=hx$ for all $x\in X$.
This means that $h^{-1}gx=x$ for all $x\in X$.
Hence $h^{-1}g$ is in the kernel of the action which is trivial since $G$ acts faithfully on $X$.
Thus $h^{-1}g=1$ and we get $g=h$.

Thanks Matt Samuel for correction.

Corrected Answer:
Let $g$ be an element in the kernel of the action.
So $gx=x=1x$ for all $x\in X$.
By the assumption that no two elements of $G$ can act the same way on $X$, we must have the kernel to be trivial and hence we conclude that $G$ acts faithfully on $X$.

The original answer is just the converse of the problem which shows that the converse of the statement is also true.

Answer 2

You have essentially proved it. No element $g$ acts the same way as the identity on an element $x$. In particular, $gx\neq x$, so $g$ does not act trivially for any nonidentity $g$, hence the action is faithful.