So far I have reached (and understood) to the point where it is proved that $\text{Gal}(E/B(\alpha))$ is a normal group in my study of theorem 2 of the article "Galois for beginners" by John Stillwell (BTW, I had to use a lemma which is not mentioned explicitly in the proof, see postscript 1). This is half of the theorem. The next line is
so it now remains to check that $\text{Gal}(B(\alpha)/B)$ is abelian.
And I understand how it is proved in the following part of the proof. However, I don't understand, how being $\text{Gal}(B(\alpha)/B)$ abelian implies, ${\rm Gal}(E/B)/{\rm Gal}(E/B(\alpha))$ is abelian (the remaining part of theorem 2).
Can anyone please explain how being $\text{Gal}(B(\alpha)/B)$ abelian implies, $\text{Gal}(E/B)/\text{Gal}(E/B(\alpha))$ is abelian? Thanks.
See the proof below-
Postscript :
1. LEMMA 2.7.3: If $\phi$ is a homomorphism of $G$ into $\bar G$ with kernel $K$, then $K $is a normal subgroup of $G$, from Topics in Algebra by I .N. Herstein .
The first rule of the proof is:
By the homomorphism theorem for groups, it suffices to find a homomorphism of ${\rm Gal}(E/B)$ with kernel ${\rm Gal}(E/B(\alpha)$ into an abelian group.
It is then proved that the map $\cdot|_{B(\alpha)}$ in ${\rm Gal}(E/B)$ has kernel ${\rm Gal}(E/B(\alpha))$.
Then he says:
[..] This also implies that $\cdot|_{B(\alpha)}$ maps ${\rm Gal}(E/B)$ into ${\rm Gal}(B(\alpha)/B)$, so it now remains to check that ${\rm Gal}(B(\alpha)/B)$ is abelian.
The homomorphism theorem (or first isomorphism theorem) says that if $\phi:G\to H$ is a group homomorphism with kernel $K$, then $G/K\cong \phi(G)$. If $H$ is abelian, then $\phi(G)$ is abelian.
In this case, we have $\cdot |_{B(\alpha)}:{\rm Gal}(E/B)\to {\rm Gal}(B(\alpha)/B)$ with kernel ${\rm Gal}(E/B(\alpha))$.