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Let $g \in A_{n}$ be a permutation whose disjoint cycle decomposition consists of odd cycles of distinct lengths, say $m$ cycles with distinct odd lengths $r_1,\cdots,r_m$.

Prove that $g$ is conjugate to $g^{-1}$ if and only if $\sum_{j=1}^{m}\frac{r_j-1}{2}$ is even.

Remark: By the spliting criterion we know the class $g^{S_n}$ splits into two equal sized classes of $g$ in $A_n$ with representatives $g$ and $(1,2)^{-1}g(1,2)$ respectively. We need to restrict $r_1,\cdots,r_m$ which forces $g^{-1}$ lies in the class of $g$.

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If $(a_1\; a_2\; \cdots\; a_{r_\nu})$ is a cycle of $g$ and $g^{-1} = h^{-1} g h$, then $(h(a_1)\; h(a_2)\;\cdots\;h(a_{r_\nu}))$ is a cycle of $g^{-1}$. Since all cycles in $g$ (and $g^{-1}$) have different lengths, we see that $(h(a_1)\; h(a_2)\;\cdots\;h(a_{r_\nu}))$ is the inverse of $(a_1\; a_2\; \cdots\; a_{r_\nu})$, hence equals $(a_k\;a_{k-1}\;\cdots\;a_1\;a_{r_\nu}\;\cdots \;a_{k+1})$ for some $k$. Hence $h(a_i)=a_j$ if $i+j\equiv k+1\pmod{r_\nu}$. Since 2 is invertible $\bmod r_\nu$, there is an $i$ with $2i\equiv k+1\pmod{r_\nu}$, hence $h(a_i)=a_i$. In all other cases, $h(a_i)=a_j$ implies also $h(a_j)=a_i$, i.e. the operation of $h$ on $\{a_1,\ldots,a_{r_i}\}$ necessarily consists of $\frac{r_i-1}2$ two-cycles. Hence $h$ consists in total of $\sum\frac{r_i-1}2$ two-cycles and from $h\in A_n$ we conclude that this sum must necessarily be even. On the other hand, we can indeed construct a $h$ as above (e.g. per cycle by letting $h(a_i)=a_{r_\nu-i}$ for $1\le i<r_\nu$) with $\sum\frac{r_i-1}2$ two-cycles and hence the condition is also sufficient.

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$g^{-1}$ has the same cycle structure as $g$. Since the cycles have distinct odd lengths, all permutations that renumber the cycle decomposition of $g^{-1}$ to that of $g$ have the same parity, so we can use the simplest one, the one that swaps opposite elements, e.g. $(12345)$ is renumbered using $(15)(24)$. This has parity $(-1)^{(r_j-1)/2}$ for a cycle of length $r_j$, so the overall parity is the sum in question.

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