any permutation times a k-cycle times the inverse is a k-cycle

**temp31415** · December 9th 2007, 02:23 PM

Question: Show that if g is a k-cycle, then $fgf^{-1}$ is also a k-cycle, for any permutation $f$ .

If the cycle decomposition of either $f$ or $f^{-1}$ was disjoint with g then they would commute and I'd be left with just $g$ which is a k-cycle but I'm not sure that $g$ is disjoint to either the cycle decomposition of $f$ or $f^{-1}$ . What am I missing?

Thanks!

**ThePerfectHacker** · December 9th 2007, 02:25 PM

Originally Posted by temp31415

Question: Show that if g is a k-cycle, then $fgf^{-1}$ is also a k-cycle, for any permutation $f$ .

If the cycle decomposition of either $f$ or $f^{-1}$ was disjoint with g then they would commute and I'd be left with just $g$ which is a k-cycle but I'm not sure that $g$ is disjoint to either the cycle decomposition of $f$ or $f^{-1}$ . What am I missing?

Thanks!

If $\sigma = (a_1,a_2,...,a_k)$ and $\tau$ is any permutation then $\tau \sigma \tau^{-1} = (\tau (a_1),...,\tau (a_k))$ .

**temp31415** · December 9th 2007, 02:33 PM

Originally Posted by ThePerfectHacker

If $\sigma = (a_1,a_2,...,a_k)$ and $\tau$ is any permutation then $\tau \sigma \tau^{-1} = (\tau (a_1),...,\tau (a_k))$ .

I don't think I understand. Could you explain it in a bit more detail?

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