Lemma 4

Tag 00IT, permanent. Cite it as [Bourbaki, Tag 00IT].

Algebra, VIII, p. 300, § 16. Other Descriptions of the Brauer Group

If $c: H\times H\rightarrow F$ is a 2-cocycle of H with values in F, then $\widetilde{c}_s$ is a 2-cocycle of G with values in F.

Let $g_1,g_2$ , and $g_3$ be elements of G. For every $i\in \{1,2,3\}$ , we define a mapping $h_i: H\backslash G\rightarrow H$ by the relation

h_i(x) =s(x\cdot g_1\cdots g_{i-1})g_is(x\cdot g_1\cdots g_i)^{-1}

for $x\in H\backslash G$ . Note that

h_1(x)h_2(x) =s(x)g_1g_2s(x\cdot g_1g_2)^{-1}

h_2(x)h_3(x) =s(x\cdot g_1)g_2g_3s(x\cdot g_1g_2g_3)^{-1}

for $x\in H\backslash G$ . We then have the relations

^{g_1}\widetilde{c}_s(g_2, g_3)\widetilde{c}_s(g_1g_2, g_3)^{-1}\widetilde{c}_s(g_1, g_2g_3)\widetilde{c}_s(g_1, g_2)^{-1}

=\prod_{x\in H\backslash G}^{g_1s(x)^{-1}}c(s(x)g_2s(x\cdot g_2)^{-1}, s(x\cdot g_2)g_3s(x\cdot g_2g_3)^{-1})

\times \prod_{x\in H\backslash G}^{s(x)^{-1}}(c(h_1(x)h_2(x), h_3(x))^{-1}

\times c(h_1(x), h_2(x)h_3(x))c(h_1(x), h_2(x))^{-1})

=\prod_{x\in H\backslash G}^{s(x\cdot g_1^{-1})^{-1}h_1(x\cdot g^{-1}_1)}c(h_2(x\cdot g_1^{-1}), h_3(x\cdot g_1^{-1}))

\times \prod_{x\in H\backslash G}^{s(x)^{-1}h_1(x)}c(h_2(x), h_3(x))^{-1}

= 1

where the second equality follows from the fact that $c$ is a 2-cocycle.

Bourbaki sets the proof directly after the statement, and the corpus does not separate them, so what is above is the statement and its proof together.

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Cites

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