# Isomorphisms and 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 $$s' \in S'$$

Last edited:
Related Linear and Abstract Algebra News on Phys.org

#### 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 [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.

### Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving