Explaining a step in the proof of the Galois group of the composite field of two Galois fields with trivial intersection Again, I am trying to understand the proof of Proposition 7 of Jyrki Lahtonen's introduction on linearly disjoint field extensions.
Explicitly, we assume that $L$ and $L'$ are both finite Galois extensions of a base field $K$. Also, let $E/K$ be a separable field which contains both $L$ and $L'$. In particular, the composite field $LL'$ is also separable over $K$. Thus the extension $LL'/K$ has a normal closure inside $E$, i.e. $N$ is the smallest Galois extension of $K$ which contains $LL'$.
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So far, I understood everything but the last line which does not seem to be too difficult. But I thought the follwing: Since $G \simeq H \times H'$, we have $\operatorname{Gal}(L/K) \simeq G/H \simeq (H \times H')/H \simeq 1 \times H' \simeq H'$ instead of $\operatorname{Gal}(L/K) \simeq H$ (and similarly for $\operatorname{Gal}(L'/K) \simeq H'$).
Could you tell me if there is a mistake in the proof or am I in the wrong? I am also not sure what is meant by the parallelogram law, so maybe clarification about that helps too.
 A: I'm not an expert, so take this with a pinch of salt, but I think there's a typo in the proof. It should say $H \simeq \mathcal{G}(L'/K)$ and $H' \simeq \mathcal{G}(L/K)$, since we have a homomorphism
$$
\theta:\mathcal{G}(N/L)\to\mathcal{G}(L'/K),\quad \sigma\mapsto\sigma\mid_{L'}
$$
given by restriction of the map $\sigma$ to the domain $L'$. 
Injectivity: This map is injective, since if $\theta(\sigma) = \text{id}_{L'}$, then $\sigma$ fixes every element of $L$ (since it is in $\mathcal{G}(N/L)$) and every element of $L'$, hence also every element of $N$, hence $\ker \theta = \{\text{id}_N\}.$
Surjectivity: The map is also surjective, because the fixed field satisfies $L'^{(\text{Im } \theta)} = K$, and hence by the Galois correspondence $\text{Im }\theta = \mathcal{G}(L'/K)$. To see that $L'^{\text{Im }\theta} = K$, let $\alpha \in L'\setminus K$, and note that since $\alpha \in N\setminus K$, there is some $\sigma \in \mathcal{G}(N/K)$ with $\sigma(\alpha) \neq \alpha$. Then $\theta(\sigma)(\alpha) = \sigma(\alpha) \neq \alpha$, and $\theta(\sigma) \in \text{Im }\theta$, so $\alpha \not \in L'^{\text{Im }\theta}$.
