# 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?

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I think it's a typo, it should say $m$ instead of $n$:

From the fact that $(\sigma \tau\sigma^{-1})^m=1$, we deduce the order must be less or equal to $m$. But it cannot be less, because if the order was less: $k$, then it would imply as they do, that $\tau^k=1$, and that's not true because the order of $\tau$ is $m>k$., and therefore the order of $\sigma\tau\sigma^{-1}$ is $m$.

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ah ok. I was a bit confused because of the typo. tnx –  MSKfdaswplwq Jun 12 '13 at 20:49

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.

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Obviously there's a typo in the given solution and you should replace $n$ by $m$.

Recall that the order of a permutation $\sigma$ ( and generally the order of an element in a group ) is the smallest positive integer $m$ s.t. $\sigma^m=1$

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Suppose that $(\sigma\tau\sigma^{-1})^k=1$. By an easy induction on $k$ we see that $(\sigma\tau\sigma^{-1})^k=\sigma\tau^k\sigma^{-1}$. Informally this is just

$$(\sigma\tau\sigma^{-1})^k=\underbrace{\underbrace{\sigma\tau\sigma^{-1}}\underbrace{\sigma\tau\sigma^{-1}}\ldots\underbrace{\sigma\tau\sigma^{-1}}}_{k}\;,$$

and the interior $\sigma^{-1}\sigma$ pairs cancel to leave just $\sigma\tau^k\sigma^{-1}$. So now we have $\sigma\tau^k\sigma^{-1}=1$. Multiply on the left by $\sigma^{-1}$ and on the right by $\sigma$:

$$\tau^k=1\tau^k1=\sigma^{-1}\sigma\tau^k\sigma^{-1}\sigma=\sigma^{-1}1\sigma=\sigma^{-1}\sigma=1\;.$$

In other words, if $(\sigma\tau\sigma^{-1})^k=1$, then $\tau^k=1$ as well. Thus, $k$ must be a multiple of the order of $\tau$, and it follows that the order of $\sigma\tau\sigma^{-1}$ is at least as big as the order of $\tau$. (In fact they are equal.) If the order of $\tau$ really is $m$, then the conclusion should be that the order of $\sigma\tau\sigma^{-1}$ is also $m$; if the conclusion is correct, that the order of $\sigma\tau\sigma^{-1}$ is $n$, then the hypothesis should have been that the order of $\tau$ is $n$, not $m$. One of the two is a typo.

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Added generality: let $G$ be a group and, given $g,h\in G$, write $g^{h}$ for $hgh^{-1}$. Then an easy induction on $n\in \mathbb{N}$ shows that, for each natural number $n$ and for each $g,h\in G$, $(g^{n})^{h}=(g^{h})^{n}$ (obviously $g^{n}$ is the $n-$th power of $g$). In particular, $g$ and $g^{h}$ has the same order for any $g,h$.