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This is the final step into completing a problem and I am a bit stuck. I need to show that:

Consider the action of $S_{n-1}$ on $S_n/S_{n-1}$ by left multiplication. Does this action have exactly two orbits? IF not, is there an action that would give exactly two orbits?

Thank you in advance!

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By $S_n/S_{n-1}$ you mean the set of left cosets of $S_n$ with respect to $S_{n-1}$? –  joriki Nov 13 '12 at 7:33
    
I mean $\{a \circ S_{n-1} \ | \ a \in S_n\}$. –  user44069 Nov 13 '12 at 7:35
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1 Answer

up vote 1 down vote accepted

Hint: The left cosets of $\pi S_{n-1}$ with $\pi\in S_n$ can be characterized by $\pi(n)$, since all elements of $\pi S_{n-1}$ send $n$ to $\pi(n)$. Using this characterization, you can see which left cosets can be transformed into each other by left-multiplication with an element of $S_{n-1}$ and which can't.

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Just a clarifying question. Is my assertion that there are exactly two cosets true? I will use your hint to prove it if this is true. –  user44069 Nov 13 '12 at 7:47
    
@Stefan: I suspect you mean orbits, not cosets? Yes, there are exactly two orbits. –  joriki Nov 13 '12 at 7:48
    
Indeed! Thank you very much! –  user44069 Nov 13 '12 at 7:50
    
@Stefan: You're welcome! –  joriki Nov 13 '12 at 7:55
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