Problem Statement
Let $\tau \in S_n$ and suppose that $\sigma = (k_1 k_2 … k_j)$ be a $j$-cycle. Prove that the conjugate of $\sigma$ by $\tau$ is also a $j$-cycle and is given by $$ \tau \sigma \tau^{-1} = (\tau(k_1) \tau(k_2) … \tau(k_j))$$
I have a proof outlined for the first part of this question, that I'm hoping is correct, but am struggling a bit to see how the second part comes together.
Proof of $1^{st}$ part
Let $\tau$ be a bijection $\tau: \{1,2,…,n\} \to \{1,2,…,n\} \in S_n$ and $\sigma = (k_1 k_2 … k_j) \in S_n$ be a $j$-cycle. Then from a lemma in the text we are guaranteed the existence of an isomorphism $c_{\tau}: S_n \to S_n$ defined by $$c_{\tau}(\sigma) = \tau \sigma \tau^{-1}$$
Suppose that $\tau$ is a transposition (a $2$-cycle) $\tau = (t_1\ t_2) = \tau^{-1}$ then for the composition $\tau \sigma \tau^{-1}$ we have $3$ separate cases:
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If $\sigma, \tau$ are disjoint cycles then they commute and we have $$ \tau \sigma \tau^{-1} = \tau (\sigma \tau^{-1}) = \tau (\tau^{-1} \sigma) = (\tau \tau^{-1}) \sigma = \sigma$$ Which is clearly a $j$-cycle.
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If $\sigma, \tau$ share one element then we have $\sigma = (k_1 … k_j)$ and $\tau = (k_f\ t)$ where $ 1 \leq f \leq j$. Then the composition $\tau \sigma \tau^{-1}$ fixes the shared element $k_f$ and we have $$ \tau \sigma \tau^{-1} = (k_1 … k_{f-1}\ t\ k_{f+1} … k_j)$$ Which is also a $j$-cycle.
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If $\sigma, \tau$ share $2$ elements then we have $\sigma = (k_1 … k_j)$ and $\tau = (k_f\ k_g)$ where, without loss of generality we have $f \lt g$. Then the composition simply transposes $k_f$ and $k_g$: $$ \tau \sigma \tau^{-1} = (k_1 … k_{f-1}\ k_g\ k_{f+1}… k_{g-1}\ k_f\ k_{g+1} … k_j)$$ Which is, again, a $j$-cycle.
Now, suppose that $\tau$ is a cycle of any length. Then we know that for $n \geq 2$ any $\tau \in S_n$ can be expressed as a composition of $2$-cycles. Thus we have $$\tau = \tau_1\ \tau_2 \cdots \tau_s$$ $$\tau^{-1} = \tau_s \cdots \tau_2\ \tau_1$$ So our composition is $$\tau \sigma \tau^{-1} = \tau_1\ \tau_2 \cdots \tau_s\ \sigma\ \tau_s \cdots \tau_2\ \tau_1$$ Now since composition of functions is associative we have: $$ \tau \sigma \tau^{-1} = \tau_1\ \tau_2 \cdots (\tau_s\ \sigma\ \tau_s) \cdots \tau_2\ \tau_1$$ Where, by the above, we see that $(\tau_s\ \sigma\ \tau_s)$ is a $j$-cycle, since $\sigma$ is a $j$-cycle. Thus $\tau_{s-1}(\tau_s\ \sigma\ \tau_s)\tau_{s-1}$ is also a $j$-cycle and consequently, by induction, $\tau \sigma \tau^{-1}$ is a $j$-cycle. $\square$
For the $2^{nd}$ part
For this part, would it be the case that since the first permutation is $\tau^{-1}$ that we can say that it operates on $\tau(i)$ for $i \in \{1,…,n\}$ so that we have $$\tau \sigma \tau^{-1}(\tau(1)\ \tau(2) … \tau(n)) = \tau \sigma (\tau^{-1}\tau(1), … , \tau^{-1}\tau(n)) = \tau \sigma(1,…,n) = \tau(k_1,…,k_j) = (\tau(k_1),…,\tau(k_j))$$
Since $\tau(i) \in \{1,…,n\}$ since $\tau \in S_n$?
Best Answer
You might be making this a bit more complicated than it need be. The cycle notation $\sigma = (k_1 \, k_2 \, \dots \, k_j)$ is just shorthand for asserting that $\sigma(k_i) = k_{i+1}$ with indices taken mod $j$, and that $\sigma$ fixes every other point.
Now, given any $\tau \in S_n$, let’s verify that it behaves similarly. Mimicking your calculation in the second part, $$ \bigl( \tau \sigma \tau^{-1} \bigr) \tau(k_i) = \tau \sigma(k_i) = \tau(k_{i+1}), $$ again with indices mod $j$. This shows that the image of $j$-cycle $\sigma$ under conjugation by $\tau$ contains a $j$-cycle.
It remains to show that the other points are fixed. Consider $k’ \in \{1, \dots, n\} \setminus \{\tau(k_1), \dots, \tau(k_j)\}$. We need to verify that $\tau \sigma \tau^{-1} (k’) = k’$. But since $\tau$ is a bijection, there is a unique $k’’ \in \{1, \dots, n\}$ such that $\tau(k’’) = k’$, namely $k’’ = \tau^{-1}(k’)$. From this definition, we can see that $k’’ \not\in \{k_1, \dots, k_j\}$, so it’s fixed by $\sigma$. Hence, $$ \bigl( \tau \sigma \tau^{-1} \bigr) (k’) = \bigl( \tau \sigma \tau^{-1} \bigr) \tau(k’’) = \tau \sigma (k’’) = \tau(k’’) = k’. $$
If you look carefully, all that we actually need is that conjugation of disjoint cycles are again disjoint cycles of the same sizes, and the fact that fixed points are just $1$-cycles. Thus, the more general statement is also true: conjugation preserves cycle shape.