Below is the proof from Ian Stewart Galois Theory 4th edition:
My question: I struggle with the texts highlighted in yellow. I assume Stewart meant that $\rho_l \rho_j^{-1}$ is the identity on $\mathbf{L}$. Indeed $\rho_l \rho_j^{-1}$ is the identity on $K(\alpha)$; but why is it the identity on $L$ as well? What does "defined by its action on $\alpha$" mean exactly? Any help would be greatly appreciated.
We have chain of subfields $K \subseteq K(\alpha) \subset L \subseteq N$. ($N$ is normal closure of $L$)
$[L:K(\alpha)]=s$ and there are $s$ embeddings of $L$ (into some nice field), which fixes $K(\alpha)$ (hence fixes $\alpha$): $$ \rho_1, \rho_2, \ldots, \rho_s: L\rightarrow N. $$ There are $r$ automorphisms of $N$ $$ \tau_1, \tau_2, \ldots, \tau_r:N\rightarrow N; $$ what these automorphisms do? $\alpha$ is root of irreducible polynomial over $K$ of degree $r$; so it has $r$ roots (in some extension) say $(\alpha=\alpha_1), \alpha_2, \ldots, \alpha_r,$ and $\tau_i$ takes $\alpha$ to $\alpha_i$.
Consider the compositions: $$ L\xrightarrow{\rho_a} N \xrightarrow{\tau_b} N\,\,\,\,\mbox{ and } \,\,\,\, L\xrightarrow{\rho_c} N \xrightarrow{\tau_d} N . $$ When they can be equal? Since $\rho$'s fix $\alpha$, so if these compositons are equal on $L$, then they are equal at $\alpha$, hence $$ \tau_b(\alpha)=\tau_d(\alpha) \Rightarrow \alpha_b=\alpha_d\Rightarrow b=d \mbox{ i.e. } \tau_b=\tau_d. $$ Then $\tau_b\circ \rho_a=\tau_b\circ \rho_b$ implies $\tau_b^{-1}\circ \tau_b=\rho_b\circ \rho_a^{-1}$ on $L$ (since left side has domain $N$ and right side has domain $L\subset N$).
So $\rho_b\circ\rho_a^{-1}$ is identity on $L$, i.e. $\rho_b=\rho_a$ on $L$. q.e.d.