# Isomorphisms and actions

1. Jun 23, 2010

### 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 $$s' \in S'$$

Last edited: Jun 23, 2010
2. Jun 23, 2010

### Martin Rattigan

You might like to have another go at that.

Strangenesses include:
Do you then mean $f$ is just a 1-1 mapping? Onto $B$?
The rôles of set and semigroup appear to have changed places.
Is the argument of the function on the last line intended to be in its domain?

3. Jun 23, 2010

### 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 [itex]s \in S[/tex] and [itex]a \in A[/tex], the expression $$f(t(a,s))=...$$ should now make sense.

I will correct also the first post in case other readers stumble upon it.