Isomorphisms and actions

  • Thread starter mnb96
  • Start date
710
5
Hello,
let's suppose we are given a set [itex]A[/itex], a (semi)group [itex]S[/itex] and we define a (semi)group-action [itex]t:A \times S \rightarrow A[/itex].
Now, if I define a bijection [itex]f:A \rightarrow B[/itex], is it possible to show that there always exists some other (semi)group S' and some action [tex]t':B \times S' \rightarrow B[/tex] such that:

[tex]\forall a \in A[/tex] and [tex]\forall s \in S[/tex]

[tex]f(t(a,s))=t'(f(a),s')[/tex]

for some [tex]s' \in S'[/tex]
 
Last edited:
You might like to have another go at that.

Strangenesses include:
... we are given a set [itex]A[/itex] ... I define an isomorphism [itex]f:A \rightarrow B[/itex]...
Do you then mean [itex]f[/itex] is just a 1-1 mapping? Onto [itex]B[/itex]?
...we define a (semi)group-action [itex]t:A \times S \rightarrow S[/itex] ... some action [tex]t':B \times S' \rightarrow B[/tex]
The rôles of set and semigroup appear to have changed places.
...[itex]f:A \rightarrow B[/itex]...[itex]t:A \times S \rightarrow S[/itex]
[tex]f(t(a,s))=\dots[/tex]
Is the argument of the function on the last line intended to be in its domain?
 
710
5
damn, I´m sorry for those mistakes. I must be very tired at this time.
btw, I´ll try to clarify:

*) [itex]f:A\rightarrow B[/itex] is a bijection

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

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

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

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