Exercise 14.2.1

Proof. Given ${\sigma }^{\prime }=({\tau }^{\prime };{\mu }_1^{\prime },...,{\mu }_k^{\prime }),\sigma =(\tau ;{\mu }_1,...,{\mu }_k)\in A\wr B$. Since ${\sigma }^{\prime }$ maps $R_i$ to $R_{{\tau }^{\prime }(i)}$ via ${\mu }_i^{\prime }$, if we set $j={\tau }^{\prime }(i)$, then $\sigma$ maps $R_j$ to $R_{\tau (j)}=R_{\tau ({\tau }^{\prime }(i))}=R_{\tau {\tau }^{\prime }(i)}$ via ${\mu }_j={\mu }_{{\tau }^{\prime }(i)}$.

Hence $\sigma {\sigma }^{\prime }$ maps $R_i$ to $R_{\tau {\tau }^{\prime }(i)}$ via ${\mu }_{{\tau }^{\prime }(i)}{\mu }_i^{\prime }$.

More explicitly, by the definition of $(\tau ;{\mu }_1,\dots ,{\mu }_k)$, for all $(i,j)\in \{1,\dots ,k\}\times \{1,\dots ,l\}$,

Applying three times this definition, we obtain

Since this equality is true for all $(i,j)\in \{1,\dots ,k\}\times \{1,\dots ,l\}$,

□