Given two permutations $g$ and $g'$ of $\{1,...,n\}$, we denote the their number of cycles by $\chi(g)$ and $\chi(g')$. Here $\chi(g)$ and $\chi(g')$ counts all cycles in $g$ And $g'$ respectively, including the cycles of length 1, e.g. the identity $I$ has cycle number $\chi(I)=n$. It seems to me that there is a bound of the cycle number for the composed permutation $g\circ g'$ as follows \begin{equation} \chi(g)+\chi(g')\leq \chi(g\circ g')+n. \end{equation} The inequality saturates at e.g. $g=I$ or $g'=I$.
I have done various experiments, and I didn't find any counter-example. But I don't know how to prove this bound. Could anyone help me to prove or disprove it? Thanks.
Let $C(g)=n-\chi(g)$. Then you are asking whether $$ C(g\circ g') \le C(g) + C(g'). $$ This is equivalent to asking whether the function $d(h,h') = C(h^{-1}h')$ is a metric on the set of $n$-permutations. And indeed, this is known to be true: $C(g)$ is the "Cayley weight" of a permutation, equivalently the minimum number of transpositions it takes to achieve the permutation $g$. I couldn't find a superb link to information about this, but here's a question with a couple of links to follow through if you want.
If you want to prove it by hand: the triangle inequality above (equivalently, your original inequality) is also saturated when $g$ and $g'$ are disjoint cycles (meaning that they don't both move any particular element). Therefore you can decompose both $g$ and $g'$ into products of disjoint cycles, and the only case you have to prove is when $g$ and $g'$ are each single cycles that overlap.