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Let $X$ be a nonempty $G$-set. Show that if $X$ is transitive (i.e., for all $x, y$ on $X$, there is a $g$ on $G$ such that $gx=y$) Then every function $f: X\to X$ is a bijection.

There is something else, $f$ holds for the following property $f(gx) = gf(x)$, $g$ on $G$ and $x$ on $X$.

We showed injective like this: Let $x, y$ be different elements on $X$, then has $X$ is transitive there is a $g$ such that $gx = y$, and $g$ is not the identity permutation, Then $f(y) = gf(x)$, hence $f(x)$ and $f(y)$ are different.

We haven't been able of showing that $f$ is onto. We tried from the definition letting $x$ be an element of $X$, then let $f(z) = y$, thus $gf(z) = gy = x = f(gz)$, thing is we think that would imply that the preimage is not unique.

Ivo Terek
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Pokemon master
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1 Answers1

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You're headed in the right direction for surjectivity, but your argument is flawed in the details. Let $y \in X$ and you must show that there exists $x \in X$ such that $y = f(x)$. Let $y'$ be an element in the image of $f$, so that there is an $x'$ such that $y' = f(x')$. By transitivity, there is a $g \in G$ such that $y = g \cdot y'$. Thus, $$y = g \cdot y' = g \cdot f(x') = f(g \cdot x'),$$ so by taking $x = g \cdot x'$, $y = f(x)$, and we have shown that $f$ is surjective.

It is not true that $f$ must be injective in general. (The problem with your attempted proof is that it is possible for $g \cdot f(x) = f(x)$ in your proof.) However, if $X$ is a finite set, then any surjective function $f : X \rightarrow X$ must be a bijection (why?).

Michael Joyce
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