Isomorphisms and Actions: Can a Bijection Preserve Group Actions?

[mnb96]

Hello,
let's suppose we are given a set A, a (semi)group S and we define a (semi)group-action t:A \times S \rightarrow A.
Now, if I define a bijection f:A \rightarrow B, is it possible to show that there always exists some other (semi)group S' and some action t':B \times S' \rightarrow B such that:

\forall a \in A and \forall s \in S

f(t(a,s))=t'(f(a),s')

for some s' \in S'

 

[Martin Rattigan]

You might like to have another go at that.

Strangenesses include:

... we are given a set A ... I define an isomorphism f:A \rightarrow B...

Do you then mean f is just a 1-1 mapping? Onto B?

...we define a (semi)group-action t:A \times S \rightarrow S ... some action t':B \times S' \rightarrow B

The rôles of set and semigroup appear to have changed places.

...f:A \rightarrow B...t:A \times S \rightarrow S
f(t(a,s))=\dots

Is the argument of the function on the last line intended to be in its domain?

 

[mnb96]

damn, I´m sorry for those mistakes. I must be very tired at this time.
btw, I´ll try to clarify:

*) f:A\rightarrow B is a bijection

**) the action of the (semi)group S onto A is t:A\times S \rightarrow A

***) if s \in S[/tex] and a \in A[/tex], the expression f(t(a,s))=... should now make sense.<br /> <br /> I will correct also the first post in case other readers stumble upon it.