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If every element of a $G$-set is left fixed by the same element $g$ of $G$, then $g$ must be the identity $e$.

I believe this to be true, but the answers say that it's false. Can anyone provide a counter-example? Thanks!

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By a $G$-set, do you mean a set on which $G$ acts? In that case, you can let the set have one element and let all elements of $G$ fix that element. – Tobias Kildetoft Jan 29 '12 at 21:41

The statement is false because the action of $G$ need not be faithful.

Remember that an action of $G$ on a set $X$ is equivalent to a group homomorphism $G\to S_X$, where $$S_X = \{f\colon X\to X\mid f\text{ is a bijection}\}.$$

The elements of $G$ that fix every element of $X$ are those that lie in the kernel of the homomorphism $G\to S_X$; we say the action is faithful if and only if the kernel is trivial, but actions need not be faithful.

For example, take your favorite nontrivial group $G$, your favorite nonempty set $X$, and make $X$ into a $G$-set by the action $g\cdot x = x$ for all $g\in G$ and all $x\in X$. This makes $X$ into a $G$-set, but for every $g\in G$ we have that every element of $X$ is left fixed by $g$.

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For concreteness, let $G$ be the group of isometries of the plane, and let $g$ be reflection in the $x$-axis. Let $S$ be the $x$-axis. Then $g(v)=v$ for every point $v$ in $S$.

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For a nontrivial example, consider the action of a group $G$ on itself by conjugation, that is, let $g \circ h=g^{-1}hg$. Now $g \circ h=h$ for all $h \in G$ means that $g$ and $h$ commute for all $h \in G$, that is $g$ is in the center $Z(G)$. And $Z(G)$ need not be just the identity element.

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