4\mathbb{Z}),\mathbb{Z})$

Current License: CC BY-SA 4.0

12 events

when toggle format	what		by	license	comment
Aug 17, 2018 at 20:07	vote	accept	Wenzhe
Aug 14, 2018 at 14:54	answer	added	Ehud Meir		timeline score: 6
May 8, 2018 at 13:43	comment	added	Derek Holt		Yes that's right. In fact $H_i(F,V)$ and $H^i(F,V)$ are zero for all nonzero $i$. This is in VI.8 of K.S. Brown's book on Cohomology of Groups, for example.
May 8, 2018 at 12:53	comment	added	YCor		To directly address Question 2, I'm pretty sure that if $F$ is a finite group and $V$ a finite $F$-module with $\|F\|$, $\|V\|$ coprime, then $H_1(F,V)=0$ although I don't have a reference.
May 8, 2018 at 12:52	comment	added	Wenzhe		@DerekHolt Thank you very much. This is very helpful.
May 8, 2018 at 12:47	comment	added	Derek Holt		You could start with en.wikipedia.org/wiki/Schur_multiplier or groupprops.subwiki.org/wiki/Schur_multiplier
May 8, 2018 at 12:42	review	Close votes
May 8, 2018 at 23:04
May 8, 2018 at 12:35	comment	added	Wenzhe		@DerekHolt I feel sorry for my naiveness, could you give me a reference for this fact?
May 8, 2018 at 12:32	comment	added	Derek Holt		But $G$ has cyclic Sylow subgroups, from which we can conclude that $H_2(G,{\mathbb Z})=0$.
May 8, 2018 at 12:31	comment	added	Wenzhe		@YCor, Thank you. What about Question 2? There is only one non-trivial action of $\mathbb{Z}/4\mathbb{Z}$ on $\mathbb{Z}/3\mathbb{Z}$. Are there any references which might be helpful?
May 8, 2018 at 12:23	comment	added	YCor		$H_1(-,\mathbf{Z})$ is the abelianization, so here the action identifies to the original one. So Question 1 has a negative answer.
May 8, 2018 at 12:00	history	asked	Wenzhe	CC BY-SA 4.0