# A probability problem in the conjugacy classes of symmetric group

Assume that $$\sigma\in S_n$$ has the cycle type $$(p,.,p,1,..,1)$$ where $$p>2$$ is a prime and the numbers of $$1$$ maybe $$0$$. If $$\sigma_1$$ and $$\sigma_2$$ are chosen uniformly in the conjugacy class of $$\sigma$$. Assume the cycle type of $$\sigma_1 \sigma_2=(k_1,k_2,.,k_l)$$ . Is there a lower bound $$B$$ such that $$\mathrm{Pr}\big(\exists j\in \{1,2,..,l\}, \text{ s.t. } k_j \bmod 2=0\big)\geq B \geq 1/T(n),$$ where $$T$$ is a polynomial.

• You are asking for the probability that $\sigma_1\sigma_2$ has even order. I conjecture that it is smallest when $\sigma$ has only one $p$-cycle and is at least $cp/n$ for some constant $c$. Proof? You want proof? Commented Feb 27, 2023 at 11:36
• yes, I need a proof Commented Feb 28, 2023 at 1:32

Let $$kp$$ be the size of the support of $$\sigma$$. Let $$1,2,3,4$$ be four points of the ground set. The probability that $$\sigma_1$$ maps $$1 \mapsto 2$$ is $$kp/n(n-1)$$, because there is a $$kp/n$$ chance that $$1$$ is in the support and conditionally a $$1/(n-1)$$ chance that $$1 \mapsto 2$$. Conditional on that, the probability that $$3$$ is in the support is $$(kp-2)/(n-2)$$ and then a $$1/(n-2)$$ chance that $$3 \mapsto 4$$ (because $$3$$ can map to anything apart from $$2$$ and $$3$$). Therefore the probability that $$\sigma_1$$ maps $$1\mapsto2$$ and $$3 \mapsto4$$ is $$kp(kp-2)/n(n-1)(n-2)^2 \ge 1/n^4$$. Likewise the probability that $$\sigma_2$$ maps $$2 \mapsto 3$$ and $$4 \mapsto 1$$ is also at least $$1/n^4$$. The probability that both these happen is at least $$1/n^8$$ and in this case $$\sigma_1\sigma_2$$ has a $$2$$-cycle.