# The smallest subgroup containing $(1, 2)$ and $(1, 2, 3.,\ldots,n)$ is $S_n$

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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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