Let $\sigma$ be an even permutation in $S_n$($\sigma \in A_n$). Assume $\sigma = \tau\sigma\tau^{-1}$ for some $\tau \in S_n$ and assume that the type of $\sigma$ consists of distinct odd integers.
Write $\sigma$ in cycle notation(including cycles of length $1$):
$\sigma = (a_1...a_{\lambda})(b_1...b_{\mu})...(c_1...c_{\nu})$
We know that
$\tau\sigma\tau^{-1} = (\tau^{-1}(a_1)...\tau^{-1}(a_{\lambda}))(\tau^{-1}(b_1)...\tau^{-1}(b_{\mu}))...(\tau^{-1}(c_1)...\tau^{-1}(c_{\nu}))$ (this is proved previously in my book as a lemma )
Since $\lambda, \mu, ..., \nu$ are odd and distinct, we have(since all cycle lengths are distinct):
$(\tau^{-1}(a_1)...\tau^{-1}(a_{\lambda})) = (a_1...a_{\lambda})$
$(\tau^{-1}(b_1)...\tau^{-1}(b_{\mu})) = (b_1...b_{\mu})$
$\ \ \ \ \ \ ... \ \ \ \ \ \ ... \ \ \ \ \ \ ... \ \ \ \ \ \ ... \ \ \ \ \ \ $
$(\tau^{-1}(c_1)...\tau^{-1}(c_{\nu})) = (c_1...c_{\nu})$
This is were I'm a bit lost. An author of my book then says that it follows that
$\tau = (a_1...a_{\lambda})^r(b_1...b_{\mu})^s...(c_1...c_{\nu})^t$
but I cannot see why it follows. I tried playing around with the equality
$(\tau^{-1}(a_1)...\tau^{-1}(a_{\lambda})) = (a_1...a_{\lambda})$
seeing what I can extract from it, but to no avail. All I know that $\tau^{-1}(a_i) \in \{a_1, ..., a_{\lambda} \}$, and if $\tau^{-1}(a_i) = a_j$, then $\tau^{-1}(a_{i+1}) = a_{j+1}$ given $i, j \leq \lambda$.