How does showing $\tau(i) = j \implies a\tau a^{-1}(a(i)) = a(j)$ demonstrate that $a\tau a^{-1}$ has the same cycle type as $\tau$?

Solution 1:

Here we have the group $G$ acting on itself by conjugation. That is, for $g \in G$ and $h \in G$ have that $\alpha : G \times G \to G$ by $\alpha(g,h) = ghg^{-1}$.

Now for a given $h \in G$. The stabilizer $G_h$ are all $g$ such that $ghg^{-1} = h$ multiplying on the right by $g$ we get that $gh = hg$. Thus, the stabilizers are exactly the centralizers.

Now let $i_1 \to i_2 \to \ldots \to i_k \to i_1$ be a cycle in the cycle decomposition of $\tau$. Then by the above $a(i_1) \to a(i_2) \to \ldots \to a(i_k) \to a(i_1)$ is a cycle in the cycle decomposition of $a\tau a^{-1}$. Thus, they have the same same cycle type.

Now to complete the proof, we have to show that if $\tau_1$ and $\tau_2$ are two permutations with the same cycle type, then there exists a permutation $a$ such that $a\tau_1 a^{-1} = \tau_2$. But this also follows from your computation (just match cycles in the cycle types and define $a$ using that).

Thus, this tells us that the orbit of a permutation is exactly all permutations with the same cycle type. Thus, we can compute the size of the by orbit by counting. Thus, we get the size of the stabilizers and hence the size of centralizer.