S_n acting on a set with more than n elements

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SUMMARY

The group S_n can act on a set with more than n elements by permuting n of those elements while leaving the remaining elements unchanged. This action can be extended to sets of 2n elements by considering two sets of n elements. Additionally, S_n has a natural action on itself, viewed as a set of n! elements, through conjugation and multiplication. Every action of a group G can be expressed as a disjoint union of transitive actions, with each orbit representing a transitive G-set.

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  • Understanding of symmetric groups, specifically S_n
  • Familiarity with group actions and orbits
  • Knowledge of conjugation in group theory
  • Basic concepts of transitive actions and quotient sets
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I'm wondering how the group S_n can act on a set with more than n elements? I'm basically only looking for some idea as to how to get started on describing such an action and how to think of that action.

Thanks
 
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S2 can act on {1,2,3} to give {2,1,3} doesn't it? *hopes not to have just made a fatal error?*
 
altcmdesc said:
I'm basically only looking for some idea as to how to get started on describing such an action and how to think of that action.
Actions can have more than one orbit.
 
i don't follow...
 
Oh, by the way...

I'm wondering how the group S_n can act on a set with more than n elements?
There is a rather obvious action of S_n on a particular set of n! elements... :wink:
 
So, essentially, S_n can act on any set of more than n elements by just permuting n of the elements and leaving the rest alone?
 
altcmdesc said:
So, essentially, S_n can act on any set of more than n elements by just permuting n of the elements and leaving the rest alone?
That's one S_n action.

However, what if you have two sets of n elements arranged into a set of 2n elements? What can you do with that?


But the hint I was trying to give earlier is that S_n has a very natural action on itself, viewed as a set of n! elements. (Can you see what that action would be?)
 
It acts on itself by conjugation, right? Does that action apply to sets with less than n! elements as well by taking an element a in the finite set and multiplying it by pap^{-1}?
 
altcmdesc said:
It acts on itself by conjugation, right?
That's one of them.

There's another one that, in some sense, is more fundamental: it acts by multiplication.

It turns out that every transitive action of a group G is isomorphic to a quotient of this one -- that is, G acting on the set G/H by multiplication, where H is a (not necessarily normal) subgroup.

And every action of a group G is a disjoint union of transitive actions -- it's the union of its orbits, and each orbit is a transitive G-set.


a in the finite set and multiplying it by pap^{-1}?
Multiplying on the right isn't defined. (although there are generalizations where right and left multiplication can happen to your set)
 

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