I am reading Allan Clark's Elements of Abstract Algebra and he states the following about transformation groups.
We note that each $g \in G$ determines a one-to-one correspondence $g: X \to X$ given by $g(x) = g * x$ whose inverse is $g^{-1}: X \to X$.
Note that $G$ is the transformation group acting on $X$ and that Clark uses "one-to-one correspondence" in place of "bijective mapping".
Next, he defines:
A transformation group $G$ acts effectively on $X$ iff for all $g, g * x = x; \forall x \in X \implies g = e$.
I actually have a question about this, too. Can anyone give a single example of a group transformation that's not effective? Please don't make it too complicated; I'm a beginner in group theory.
The main confusion is here:
Show that the set $A(X)$ of one-to-one correspondences forms the largest effective group on $X$.
Didn't we say earlier that every transformation group is isomorphic to a set of one-to-one correspondences? How come now we're restricting ourselves to effective groups?
Consider the trivial group action, defined by $g\cdot x=x$ for every $x\in X$. It's not effective.
For a less trivial example, consider a non-trivial normal subgroup $H\triangleleft G$. Let $G$ act on the group $G/H$, by left multiplication. This won't be an effective action. Any $h\in H$ fixes each coset.
Next, $A(x)=\rm{Sym}(X)$ is the symmetric group on $X$. It consists in all the bijective maps from $X$ to itself.
A group action gives a homomorphism from $G$ to $\rm{Sym}(X)$. If the action is effective (often called faithful) it's injective. Thus the largest group acting effectively on $X$ is $A(X)=\rm{Sym}(X)$.
Any group acts effectively on itself by left multiplication (or right multiplication by the inverse).