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At http://planetmath.org/simplicityofthealternatinggroups it states the following.

Let $\pi$ be a permutation written as disjoint cycles \[ \pi = (a_1, a_2, \ldots, a_k)(b_1, b_2, \ldots, b_l)\ldots (c_1,\ldots, c_m) \] It is easy to check that for every permutation $\sigma \in S_n$ we have \[ \sigma \pi \sigma^{-1} = (\sigma(a_1), \sigma(a_2), \ldots, \sigma(a_k))\, (\sigma(b_1),\sigma(b_2), \ldots \sigma(b_l))\, \ldots (\sigma(c_1),\ldots, \sigma(c_m)) \]

While it seems clear that this is true based on trying several concrete cases, I'm having trouble figuring out how I would check this. Some guidance on this would be much appreciated.

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Applying on $\sigma(a_i)$ suppose $(\sigma \pi (\sigma)^{-1})\sigma(a_i)= \sigma \pi(a_i)=\sigma(a_{i+1})$. Now you are clear I think.

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  • $\begingroup$ Can you elaborate on the strategy here? I'm know it's probably quite simple, but bear with me, I'm very new to this. $\endgroup$ – Gerber Oct 12 '14 at 19:22
  • $\begingroup$ Never mind! It suddenly clicked! $\endgroup$ – Gerber Oct 12 '14 at 20:52

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