For elements $a,b \in G$, group that acts on the set $X$, show that $a$ and $b^{-1}ab$ have the same numbers of fixed points in $X$

Question

The group $G$ acts on the finite set $X$.

Let $a,b \in G$.

Show that $a$ and $b^{-1}ab$ have the same number of fixed points in $X$

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My idea was perhaps to define two sets which contain all the fixed points of $a$ and $b^{-1}ab$ respectively and then proceed to find a bijection between those groups. The function I thought of is defined by $f(x)=b^{-1}x$.

Would that be a sufficient proof and if so how exactly can I do that (for example is it necessary to prove it is well-defined)?

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Answer 1

\begin{alignat}{1} \operatorname{Fix}(b^{-1}ab) &= \{x\in X\mid (b^{-1}ab)\cdot x=x\} \\ &= \{x\in X\mid b^{-1}\cdot((ab)\cdot x)=x\} \\ &= \{x\in X\mid (ab)\cdot x=b\cdot x\} \\ &= \{x\in X\mid a\cdot (b\cdot x)=b\cdot x\} \\ \end{alignat}

The map $f_b\colon X\to X$, defined by $x\mapsto b\cdot x$, is bijective. Therefore:

\begin{alignat}{1} |\operatorname{Fix}(b^{-1}ab)| &= |\{x\in X\mid a\cdot (b\cdot x)=b\cdot x\}| \\ &= |\{b\cdot x\in X\mid a\cdot (b\cdot x)=b\cdot x\}| \\ &= |\{y\in X\mid a\cdot y=y\}| \\ &= |\operatorname{Fix}(a)| \\ \end{alignat}

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Following your idea, we have the two sets $A:=\operatorname{Fix}(b^{-1}ab)=\{x\in X\mid a\cdot (b\cdot x)=b\cdot x\}$ (see above) and $B:=\operatorname{Fix}(a)=\{y\in X\mid a\cdot y=y\}$; define the map $f\colon A\to B$ by $x\mapsto y:=b\cdot x$.

• Good definition: $x\in A \Rightarrow a\cdot(b\cdot x)=b\cdot x \Rightarrow a\cdot y=y \Rightarrow y\in B$;
• Injectivity:

\begin{alignat}{1} f(x)=f(x') &\Rightarrow b\cdot x=b\cdot x' \\ &\Rightarrow b^{-1}(b\cdot x)=b^{-1}(b\cdot x') \\ &\Rightarrow (b^{-1}b)\cdot x=(b^{-1}b)\cdot x' \\ &\Rightarrow e\cdot x=e\cdot x' \\ &\Rightarrow x=x' \\ \end{alignat}

• Surjectivity: $\forall y \in B, \exists x\in A\mid y=b\cdot x \iff \forall y \in B, \space x=b^{-1}\cdot y\in A$; but indeed $a\cdot(b\cdot x)=a\cdot(b\cdot (b^{-1}\cdot y))=a\cdot y\stackrel{y\in B}{=}y=b\cdot x$, and then $x\in A$.

Actually, the finiteness of $X$ doesn't seem to be relevant here.

Answer 2

Hint: define a map $\phi : Fix_X(a) \rightarrow Fix_X(b^{-1}ab)$ by $\phi(x)=x^b$. Show that $\phi$ is well-defined and both injective and surjective.