Question

Show that the inverse of $\left(a_1{a}_2......a_k\right)$ in S_n is $\left(a_k{a}_{k-1}.....a_3{a}_2{a}_1\right)$.

Short answer

The inverse of $\left({\mathrm{a}}_1{\mathrm{a}}_2......{\mathrm{a}}_{\mathrm{k}}\right)$in S_n is $\left({\mathrm{a}}_{\mathrm{k}}{\mathrm{a}}_{\mathrm{k}-1}....{\mathrm{a}}_3{\mathrm{a}}_2{\mathrm{a}}_1\right)$.

Step-by-step solution

Required Theorem

Theorem 7.26: Every permutation in S_n is a product of (not necessarily disjoint) transpositions.

Proving that the inverse of (a1a2....ak) in Sn is (akak-1.....a3a2a1).

From the above theorem, we can write both the given cycle as a product of its transpositions.

$\left({\mathrm{a}}_1{\mathrm{a}}_2....{\mathrm{a}}_{\mathrm{k}}\right)=\left({\mathrm{a}}_1{\mathrm{a}}_2\right)\left({\mathrm{a}}_2{\mathrm{a}}_3\right).....\left({\mathrm{a}}_{\mathrm{k}}{\mathrm{a}}_{\mathrm{k}-1}\right)$

and

$\left({\mathrm{a}}_{\mathrm{k}}{\mathrm{a}}_{\mathrm{k}-1}.....{\mathrm{a}}_3{\mathrm{a}}_2{\mathrm{a}}_1\right)=\left({\mathrm{a}}_{\mathrm{k}}{\mathrm{a}}_{\mathrm{k}-1}\right)\left({\mathrm{a}}_{\mathrm{k}-1}{\mathrm{a}}_{\mathrm{k}-2}\right).........\left({\mathrm{a}}_3{\mathrm{a}}_2\right)\left({\mathrm{a}}_2{\mathrm{a}}_1\right)$

So,

$\left({\mathrm{a}}_1{\mathrm{a}}_2......{\mathrm{a}}_{\mathrm{k}}\right)\left({\mathrm{a}}_{\mathrm{k}}{\mathrm{a}}_{\mathrm{k}-1}....{\mathrm{a}}_3{\mathrm{a}}_2{\mathrm{a}}_1\right)=\left({\mathrm{a}}_1{\mathrm{a}}_2\right).\left({\mathrm{a}}_2{\mathrm{a}}_3\right)......\left({\mathrm{a}}_{\mathrm{k}-1}{\mathrm{a}}_{\mathrm{k}}\right).\left({\mathrm{a}}_{\mathrm{k}}{\mathrm{a}}_{\mathrm{k}-1}\right).\left({\mathrm{a}}_{\mathrm{k}-1}{\mathrm{a}}_{\mathrm{k}-2}\right).......\left({\mathrm{a}}_3{\mathrm{a}}_2\right).\left({\mathrm{a}}_2{\mathrm{a}}_1\right)$

We know, $\left({\mathrm{a}}_{\mathrm{k}-1}{\mathrm{a}}_{\mathrm{k}}\right).\left({\mathrm{a}}_{\mathrm{k}}{\mathrm{a}}_{\mathrm{k}-1}\right)$=1,

$\left({\mathrm{a}}_1{\mathrm{a}}_2......{\mathrm{a}}_{\mathrm{k}}\right)\left({\mathrm{a}}_{\mathrm{k}}{\mathrm{a}}_{\mathrm{k}-1}....{\mathrm{a}}_3{\mathrm{a}}_2{\mathrm{a}}_1\right)=\left({\mathrm{a}}_1{\mathrm{a}}_2\right).\left({\mathrm{a}}_2{\mathrm{a}}_3\right)....\left({\mathrm{a}}_{\mathrm{k}-2}{\mathrm{a}}_{\mathrm{k}-1}\right).\left({\mathrm{a}}_{\mathrm{k}-1}{\mathrm{a}}_{\mathrm{k}-2}\right)....\left({\mathrm{a}}_3{\mathrm{a}}_2\right)\left({\mathrm{a}}_2{\mathrm{a}}_1\right)$.

Again, $\left({\mathrm{a}}_{\mathrm{k}-2}{\mathrm{a}}_{\mathrm{k}-1}\right).\left({\mathrm{a}}_{\mathrm{k}-1}{\mathrm{a}}_{\mathrm{k}-2}\right)$=1

So, the last the equation will be:

$\left({\mathrm{a}}_1{\mathrm{a}}_2.....{\mathrm{a}}_{\mathrm{k}}\right).\left({\mathrm{a}}_{\mathrm{k}}{\mathrm{a}}_{\mathrm{k}-1}....{\mathrm{a}}_3{\mathrm{a}}_2{\mathrm{a}}_1\right)=1$

Since the product of both is 1. Hence, it is proved that the inverse of $\left({\mathrm{a}}_1{\mathrm{a}}_2.....{\mathrm{a}}_{\mathrm{k}}\right)$in S_n is $\left({\mathrm{a}}_{\mathrm{k}}{\mathrm{a}}_{\mathrm{k}-1}.....{\mathrm{a}}_3{\mathrm{a}}_2{\mathrm{a}}_1\right)$.