Let $H\leq G=\operatorname{Gal}(K/F)$ ($K/F$ is a finite galois extension), why is the following map well defined:

$\varphi:G/H\to\Gamma_F(K^H,K)$ defined by $\sigma H\mapsto\sigma|_{K^H}$ ,where $\Gamma_{F}(K^H,K)$ denotes all homomorphisms from $K^H$ to $K$ that fixes $F$.

My lecture wrote : Let $\sigma\in G$ ,if $\tau\in H$ then $\tau|_{K^H}=\operatorname{Id}_{K^H}$hence $\sigma\tau|_{K^H}=\sigma|_{K^H}$. Why does this imply (what I understand that need to be shown): $\sigma_1|_{K^H}=\sigma_2|_{K^H}\implies\sigma_2^{-1}\sigma_1\in H$ ?

Please note I was not told $H$ is a normal subgroup of $G$.

  • $\begingroup$ Note that if $H$ is just an arbitrary subgruop, then $G/H$ represents the left cosets of $H$ in $G$, rather than a quotient. $\endgroup$ – Arturo Magidin Jun 8 '12 at 17:23
  • $\begingroup$ @ArturoMagidin - I noted this in the question, do you think that what I wrote regarding how I understand the meaning here to be well defined is wrong ? (I tried being carefull) $\endgroup$ – Belgi Jun 8 '12 at 17:25
  • $\begingroup$ Well, you never said that $G/H$ was the cosets, and then you make an apparently big deal of noting that $H$ is not assumed to be normal, as if that was part of your confusion. Hence the comment. $\endgroup$ – Arturo Magidin Jun 8 '12 at 17:27
  • $\begingroup$ @ArturoMagidin I noted this because when I tried to prove it I got to the case that if I could say $H$ is normal then I would of been done. sorry if there was comfusion (isn't $G/H$ the standard notaion even if $H$ is not normal ?) $\endgroup$ – Belgi Jun 8 '12 at 17:30
  • $\begingroup$ Sometimes, in some books. In others, it is reserved for the quotient group. $\endgroup$ – Arturo Magidin Jun 8 '12 at 18:19

Note that $G/H$ represents the left cosets of $H$ in $G$.

Let $\sigma\in G$. Then for every $\tau\in H$ we have that the value of $\sigma$ and of $\sigma\tau$ on $K^H$, the field fixed by $H$, is the same. Thus, every element of $\sigma H$ determines the same map $K^H\to K$. This means that the map $G\to\mathrm{Hom}_F(K^H,K)$ given by restriction actually factors through the cosets.

Moreover, if $\sigma_1,\sigma_2\in G$ are such that $\sigma_1|_{K^H} = \sigma_2|_{K^H}$, then for every $a\in K^H$ we have $\sigma_1(a)=\sigma_2(a)$, hence $\sigma_2^{-1}\sigma_1(a) = a$. Thus, $\sigma_2^{-1}\sigma_1$ fixes every $a\in K^H$, hence lies in $H$ (since the extension is Galois, the stabilizer of $K^H$ is exactly $H$). Thus, $\sigma_1H = \sigma_2H$. Hence, the restriction map factors through the cosets but not through any larger subgroup.

  • $\begingroup$ I believe you should restrict $\sigma_i$ to $K^H$ and not $H$. Thank you for your time and help! $\endgroup$ – Belgi Jun 8 '12 at 17:35
  • $\begingroup$ @Belgi: Indeed I should. Fixed. $\endgroup$ – Arturo Magidin Jun 8 '12 at 18:18

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