$S_n$ is the set of all permutations.
Show that $(στ)^{-1} = τ^{-1}σ^{-1}$ for all $σ, τ ∈ S_n$
I can somewhat see why this statement would be true, seeing as permutations are read from right to left which could have something to do with the order switch. I'm just not sure there are many options to actually show that the equality is true.
For any two elements $a,b$ in $S_n$ we have $(ab)^{-1}=b^{-1}a^{-1}$.
That is because $$(ab)(b^{-1}a^{-1})=a(bb^{-1})a^{-1}=aa^{-1}=1$$ and similarly $$(b^{-1}a^{-1})(ab)=b^{-1}(a^{-1}a)b=b^{-1}b=1.$$
We are using associativity property of compositions.
Recall that $\sigma^{-1}$ is defined by the rule $\sigma\sigma^{-1}=Id=\sigma^{-1}\sigma$. Hence we need to check that $(\sigma\tau)(\tau^{-1}\sigma^{-1})=Id=(\tau^{-1}\sigma^{-1})(\sigma\tau)$. But this is easy since $(\sigma\tau)(\tau^{-1}\sigma^{-1})=\sigma(\tau\tau^{-1})\sigma^{-1}=\sigma \circ Id \circ \sigma^{-1}=\sigma\sigma^{-1}=Id. $
