# Let $a_i$ and $a_j$ disjoint cycles of $S_n$. Since $τ∈S_n$ is an one-to-one function, $τα_i τ^{-1}$ and $τα_jτ^{-1}$ must be also disjoint.

My solution manual says that:

Let $a_i$ and $a_j$ disjoint cycles of $S_n$. Since $τ∈S_n$ is an one-to-one function, $τα_i τ^{-1}$ and $τα_jτ^{-1}$ must be also disjoint.

This make sense to me, but I'm trying to prove this more rigoursly to myself. How could I do this ?

Second try:
Suppose $τα_i τ^{-1}$ doesn't fix $i$ and suppose $τα_jτ^{-1}$ doesn't fix $j$.
Then I need to prove that $i≠j$.
So I know that $τα_i τ^{-1}(i)≠i$ and $τα_jτ^{-1}(j)≠j$.
As $τ^{-1}$ is injective, $α_i τ^{-1}(i)≠τ^{-1}(i)$ and $α_jτ^{-1}(j)≠τ^{-1}(j)$.
Therefore $α_i (τ^{-1}(i))≠τ^{-1}(i)$ and $α_j(τ^{-1}(j))≠τ^{-1}(j)$.
As $α_i$ and $α_j$ are disjoint: $τ^{-1}(i)≠τ^{-1}(j)$.
As $τ$ is injective: $i≠j$. Exactly what I needed to prove.

Is this a correct proof ? Could I prove this more easily ?

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+1 for a question seeking deeper understanding. –  Ben Blum-Smith Apr 12 '13 at 18:25

Yes, your proof is correct. I think it is the most straightforward way to prove this. Notice that the proof works even when $a_i$ and $a_j$ are not cycles.
Hint: Recall that $\tau (x_1 \ldots x_n) \tau^{-1} = (\tau(x_1) \ldots \tau(x_n))$.