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We are given a permutation group G acting on a finite set X. Finite set $\Omega$ contains all mappings $\omega$ from the set X to finite set K. We define mapping $\hat{g}$ on the set $\Omega$ in the following way

$(\hat{g}(\omega))(x)= \omega(g(x))$

Question: Is the statement below true?

If a,b are in G then $((\hat{a}\hat{b})(\omega))(x)=((\widehat{ab}(\omega))(x)$

I will show that it is not true:

$((\widehat{ab}(\omega))(x)=\omega((ab)(x))=\omega(a(b(x)))$

$((\hat{a}\hat{b})(\omega))(x)=(\hat{a}(\hat{b}(\omega)))(x)=(\hat{a}(\omega_p))(x)=\omega_p(a(x))=\omega(b(a(x)))$

$\hat{b}(\omega)=\omega_p$ and $(\hat{b}(\omega))(x)=\omega_p(x)=\omega(b(x))$

Are my deductions correct? If they are correct then mapping which is defined above cannot be a homeomorphism isomorphism?

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"Homoemorphism" is the wrong word, there are no topologies involved here. If you change you definition of $\hat{g}$ to $\hat{g}(\omega) :=x\mapsto\omega(g^{-1}x)$ then everything works out and you get indeed a group action of $G$ on $\Omega$. Also: This is not the right place to ask this sort of question. MO is for research level questions. Try math.stackexchange.com for example. –  Johannes Hahn Mar 9 '13 at 18:45

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The "right" way to make $G$ act on $\Omega$ is $\hat g(\omega)(x)=\omega(g^{-1}(x))$. Not only does this give you a left action (rather than a right action), but it fits nicely with the "set of ordered pairs" view of functions. If you regard $\omega$ as set of ordered pairs $(x,\omega(x))$, then the action of any $g\in G$ amounts to applying $g$ to the only place where it makes sense, namely the first components of the ordered pairs.

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In case $G=S$ is a semigroup, one can still save the ``left'' action by $\tilde{h}(\omega)(x):=(\omega(xh)$. In case $K$ is a vector space (the theory becomes then interesting when $K$ is infinite), this gives the theory of representative functions. See theorem 2.2.6 in Eichii Abe's book Hopf Algebras (Cambridge University Press). –  Duchamp Gérard H. E. Mar 9 '13 at 23:31
    
Yes, in the case of semigroups, a right action of the semigroup on $X$ will give you a left action on $\Omega$ (as in Gérard's comment), and vice versa (as in Boris's question). Of course if the semigroup itself is $X$, then you have actions on both sides. But I see no way to get, as Boris seems to want, a left action on $\Omega$ from a left action on $X$ unless inverses (or adjoints or something else that reverses the order of multiplication) are available. –  Andreas Blass Mar 10 '13 at 2:14

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