Prove that $\sigma = \tau =e $ If $\sigma , \tau $ are two permuations that disturb no common element and $\sigma \tau = e$ , prove that $\sigma = \tau =e $

Suppose that $\sigma(k)\
e k$. By hypothesis $\tau(k)=k$, so $(\sigma\tau)(k)=\sigma\big(\tau(k)\big)=\sigma(k)\
e k$, and $\sigma\tau\
e e$.

**Added:** You can even get $\tau=e$ without knowing that $\sigma$ and $\tau$ commute. Suppose that $\tau(k)\
e k$; say $\tau(k)=\ell$. Then clearly $\tau(\ell)\
e\ell$, so $\sigma(\ell)=\ell$. Thus, $(\sigma\tau)(k)=\sigma\big(\tau(k)\big)=\sigma(\ell)=\ell\
e k$, and again $\sigma\tau\
e e$.