Suppose that $g,h$ are $p$-cycles on $\Omega$ where $| \Omega| = p$, a prime. Let $G := \langle g,h \rangle$. Since $\langle g \rangle$ and $\langle h \rangle$ are of order $p$ and the maximum group on $\Omega$, $S_p$ has order $p!$, then $\langle g \rangle, \langle h \rangle \in \operatorname{Syl}_p(G)$. By Sylow's second theorem then, we get $\langle g \rangle^s = \langle h \rangle$ for some $s \in G$. Therefore, $g^s = h^k$ for some $k \in \mathbb{N}$. Now, if $s = g\hat{s}$ for some $\hat{s} \in G$, then $h^k = g^s = g^{g\hat{s}} = g^{\hat{s}}$, and if $s=h\hat{s}$, then $g^s = g^{h\hat{s}} = h^{-1}g^{\hat{s}}h = h^k \implies g^{\hat{s}} = hh^kh^{-1} = h^k$. Similarly, if $s=g^{-1}\hat{s}$ or $s=h^{-1}\hat{s}$ we get the same result. Therefore, since $s$ is a finite product of $g$, $h$, and their inverses, we can cancel it down to find $g=h^k$.
My question is: What is the problem with this argument? Obviously this can not be correct since $g$ and $h$ are arbitrary $p$-cycles on $\Omega$.
The equation $g^{h\hat{s}}=h^{-1}g^{\hat{s}}h$ is wrong, since $g^{h\hat{s}}=\hat{s}^{-1}h^{-1}gh\hat{s}$ with the $h$'s on the inside instead of the $\hat{s}$'s. To remove an $h$ from $s$ in this way it would need to be on the right instead of on the left. You could try cancelling the $g$'s and $h$'s from the right side of $s$ instead of the left, but then cancelling $g$'s would not work for a similar reason.