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

• Dec 9th 2007, 02:23 PM
temp31415
any permutation times a k-cycle times the inverse is a k-cycle
Question: Show that if g is a k-cycle, then $\displaystyle fgf^{-1}$ is also a k-cycle, for any permutation $\displaystyle f$.

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

Thanks!
• Dec 9th 2007, 02:25 PM
ThePerfectHacker
Quote:

Originally Posted by temp31415
Question: Show that if g is a k-cycle, then $\displaystyle fgf^{-1}$ is also a k-cycle, for any permutation $\displaystyle f$.

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

Thanks!

If $\displaystyle \sigma = (a_1,a_2,...,a_k)$ and $\displaystyle \tau$ is any permutation then $\displaystyle \tau \sigma \tau^{-1} = (\tau (a_1),...,\tau (a_k))$.
• Dec 9th 2007, 02:33 PM
temp31415
Quote:

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

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