Let $\gamma$ be any cycle of length $r$. If $\sigma\in S_n$, show that $\sigma\gamma\sigma^{-1}$ is also a cycle of length $r$. More precisely, if $\gamma=\begin{pmatrix} k_1 & k_2 & \cdots & k_r \end{pmatrix}$ show that $\sigma\gamma\sigma^{-1}=\begin{pmatrix} \sigma k_1 & \sigma k_2 & \cdots & \sigma k_r \end{pmatrix}$.

Let $p$ be the length of the cycle $\sigma$.

Let $\sigma=\begin{pmatrix} m_1 & m_2 & \dots & m_p\\ \end{pmatrix}$ and $\sigma^{-1}=\begin{pmatrix} n_1 & n_2 & \dots & n_p\\ \end{pmatrix}$

So I figure I can break this into 3 cases: $p=r, p>r,$ and $p<r$.

I am attempting $p=r$ first. So $\sigma\gamma\sigma^{-1}= \begin{pmatrix} m_1 & m_2 & \dots & m_r\\ \end{pmatrix}\begin{pmatrix} k_1 & k_2 & \dots & k_r\\ \end{pmatrix}\begin{pmatrix} n_1 & n_2 & \dots & n_r\\ \end{pmatrix}$

I personally find it easier to multiply with the following notation rather than cycles

$\begin{pmatrix} 1 & 2 & \dots & r\\ \sigma 1 & \sigma 2 & \dots & \sigma r \end{pmatrix}\begin{pmatrix} 1 & 2 & \dots & r\\ \gamma 1 & \gamma 2 & \dots & \gamma r \end{pmatrix}\begin{pmatrix} 1 & 2 & \dots & r\\ \sigma^{-1} 1 & \sigma^{-1} 2 & \dots & \sigma^{-1} r \end{pmatrix}$

but I am stuck at this point... Am I right to break it up into cases? I also don't see how I can multiply this out to see that $\sigma\gamma\sigma^{-1}=\begin{pmatrix} \sigma k_1 & \sigma k_2 & \cdots & \sigma k_r \end{pmatrix}$.

  • $\begingroup$ σ is not necessarily a cycle. You just have to check the formula they give you. $\endgroup$ – Bernard Oct 6 '15 at 0:24
  • $\begingroup$ In my book $S_n$ is the set of all permutations of a symmetric group of degree $n$. So if $\sigma\in S_n$, then $\sigma$ is a permutation and a permutation is a product of cycles. So I thought $\sigma\in S_n$ implies $\sigma$ is a cycle? $\endgroup$ – Burgundy Oct 6 '15 at 0:31
  • $\begingroup$ Not at all. A product of two disjoint transpositions, for instance, is never a cycle. $\endgroup$ – Bernard Oct 6 '15 at 0:33
  • $\begingroup$ So the products of cycles aren't necessarily cycles? $\endgroup$ – Burgundy Oct 6 '15 at 0:37
  • $\begingroup$ If they're disjoint they can't be. $\endgroup$ – Bernard Oct 6 '15 at 0:48

The trick is to not look at everything, the only elements of $\{1,2,\dots,n\}$ we are interested in are:

$\sigma(k_j)$, for $j = 1,2,\dots,r$.

But, you might say, what about all the OTHER elements of $\{1,2,\dots,n\}$?

Well if $a \not \in \{k_1,k_2,\dots,k_r\}$, then $\sigma(a)$ maps to something not in $\{\sigma(k_1),\sigma(k_2),\dots,\sigma(k_r)\}$ (this is because permutations are bijective).

Let's say $\sigma(a) = b$. By our choice of $b$ (that is, as $\sigma(a)$), this is not in $\{\sigma(k_1),\sigma(k_2),\dots,\sigma(k_r)\}$. Then $\sigma^{-1}(b) = a$. Since $a \not \in \{k_1,k_2,\dots,k_r\}$, the $r$-cycle $\gamma = (k_1\ k_2\ \cdots\ k_r)$ doesn't affect $a$.

So $\gamma\sigma^{-1}(b) = \gamma(a) = a$. Therefore, $\sigma\gamma\sigma^{-1}(b) =\sigma(a) = b$, so elements not in $\{\sigma(k_1),\sigma(k_2),\dots,\sigma(k_r)\}$ are not affected at ALL, by $\sigma\gamma\sigma^{-1}$.

So let's look at what happens to $\sigma(k_j)$. $j = r$ is a special case, so we'll do that first.

Well, $\sigma^{-1}(\sigma(k_r)) = k_r$ (duh!). And $\gamma(k_r) = k_1$. Finally, $\sigma(k_1) = \sigma(k_1)$ (that part was easy, heh). Putting this all together we have:

$\sigma\gamma\sigma^{-1}(\sigma(k_r)) = \sigma(k_1)$

Ok, now "all the other cases" (all at once, because we're just that good).

If $j = 1,2,\dots,r-1$, then:

$\sigma\gamma\sigma^{-1}(\sigma(k_j)) = (\sigma\gamma)(k_j) = \sigma(\gamma(k_j)) = \sigma(k_{j+1})$.

Evidently, then, $\sigma\gamma\sigma^{-1}$ is the permutation that leaves everything outside of $\{\sigma(k_1),\sigma(k_2),\dots,\sigma(k_r)\}$ fixed, and sends:

$\sigma(k_1) \mapsto \sigma(k_2)\\ \sigma(k_2) \mapsto \sigma(k_3)\\ \vdots\\ \sigma(k_r) \mapsto \sigma(k_1)$

which is the $r$-cycle we set out to prove it was.


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