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?



A yes no question concerning induced group 
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.


