Order of Permutation : If $\tau \in S_n$ has order $m$, then $\sigma \tau\sigma^{-1}$ has also order $m$. I dont understand the following very simple statement:
If $\tau \in S_n$ has order $m$, then $\sigma \tau\sigma^{-1}$ has also order $m$.
The proof is:
Suppose $\tau$ has order $m$. 
$(\sigma \tau \sigma^{-1})^{m} = \sigma \tau^{m}\sigma^{-1}= (1)$
Then (using http://www.math.niu.edu/~beachy/abstract_algebra_2ed/guide/soln23.pdf exercise 17) they say that the order of $\sigma\tau\sigma^{-1}$ cannot be less than $n$ (remember we are in $S_n$), since $(\sigma\tau\sigma^{-1})^k=1 $ implies $\sigma\tau^k\sigma^{-1} =(1)$, and then $\tau^k=\sigma^{-1} \sigma = (1)$
I don't understand the last part.... Who can help me?

One way to look at this is the following. The calculation shows that the order of $\sigma\tau\sigma^{-1}$ cannot be higher than the order of $\tau$. Write $\tau'=\sigma\tau\sigma^{-1}$ and $\sigma'=\sigma^{-1}$. The same general principle then implies that the order of $\sigma'\tau'\sigma'^{-1}$ cannot be higher than the order of $\tau'$. But
$$
\sigma'\tau'\sigma'{-1}=\sigma^{-1}(\sigma\tau\sigma^{-1})\sigma=\tau,
$$
so this is exactly the reverse inequality.
