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I try this by following way, Let $H $ be the subgroup generated by $(1, 2), (1, 2, 3,\dots .,n) $. How do I show, $H$ contain elements $ (1, r) $ for $ r = 1, 2,...n$. Does there exist any trick to show it?

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1 Answer 1

up vote 8 down vote accepted

Let $s = (1\;2)$ and $c = (1\;2\;3\;\ldots\;n)$ then $$c^{-i} s c^{i} = (i\;i+1)$$ and since you can create any permutation from transpositions, this gives the whole of $S_n$.

To verify this identity, see that $j$ gets mapped to $j-i \mod n$ then swapped only if it's 1 or 2, then $i$ is added back.


Also $c^{-(n-2)} s c^{n-1} = (1\;2\;3\;\ldots\;n-1)$


If we use $a^b$ short for $b^{-1} a b$ (the reason this is such a useful operation to do is that it preserves the length of the cycle) then consider

  • $s$
  • $(s^c)^s$
  • $(((s^c)^s)^c)^s$
  • ...
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Nice work + 1...you certainly beat me too it! –  amWhy Feb 20 '13 at 14:16

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