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It's clear $1$ is a projective object in the category of finite groups. Are there any others?

Note that the dual problem, for injective objects, is comparatively easy (using Cayley's theorem).

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  • $\begingroup$ I don't know the answer yet, I'm thinking about it; I found that such a $P$ was non abelian, and then improved the result to prove that $P$ was perfect. $\endgroup$ Jan 31, 2019 at 15:25
  • $\begingroup$ @Max Indeed, Suppose $P$ has a nontrivial abelian quotient. Then its ha a non-trivial cyclic quotient $\Bbb Z/n\Bbb Z$. But that surejction $P\to \Bbb Z/n\Bbb Z$ cannot be lifted to $\Bbb Z/nm\Bbb Z$ for $m\gg 1$: some element of $P$ would have to be mapped to a generator. $\endgroup$ Feb 3, 2019 at 16:09
  • $\begingroup$ @HagenvonEitzen : yup; I was trying some other things like using the Cayley embedding, but I don't know any nonsplit surjection $K\to \mathfrak{S}_n$; or I tried to see if there was any reason why we could find a finite quotient of the free group on $P$ generators that was larger than $P$ but no luck there. $\endgroup$ Feb 3, 2019 at 17:13

1 Answer 1

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There are no others. Here's a proof using well-known facts about the cohomology of finite groups.

Let $G$ be a nontrivial finite group, and let $p$ be a prime dividing the order of $G$. Then there is a finite dimensional (and hence finite) $\mathbb{F}_pG$-module $M$ with $H^2(G,M)\neq0$. A nonzero cohomology class represents a non-split extension $1\to M\to\tilde{G}\to G\to1$.

If the existence of $M$ is not familiar, but you believe that $H^k(G,\mathbb{F}_p)\neq0$ for some $k>0$, then you can produce $M$ from the trivial $\mathbb{F}_pG$-module $\mathbb{F}_p$ by dimension shifting. Or for a specific $M$, take an exact sequence $$0\to M\to P_1\to P_0\to\mathbb{F}_p\to0,$$ where $P_1\to P_0\to\mathbb{F}_p\to0$ is a projective presentation of the trivial $\mathbb{F}_pG$-module.

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    $\begingroup$ group cohomology question -- What if there is a splitting of the extension as abstract groups which is incompatible with the given action of $G$ on $M$? $\endgroup$
    – hunter
    Feb 4, 2019 at 11:32
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    $\begingroup$ @hunter I'm not sure what you mean by a splitting being compatible with the action on $M$. A splitting is simply a group homomorphism $G\to\tilde{G}$ such that the composition $G\to\tilde{G}\to G$ is the identity map. In any case, the action of $G$ on $M$ is determined by the map $\tilde{G}\to G$. $\endgroup$ Feb 4, 2019 at 11:37
  • $\begingroup$ Yes, but @hunter is right in that $H^2(G,M)$ assumes a given action on $M$, and it classifies extensions $1\to M\to A\to G\to 1$ such that the induced action is the given one $\endgroup$ Feb 4, 2019 at 15:44
  • $\begingroup$ @hunter : ah actually I just realized it doesn't make sense to ask that : a nontrivial cohomology class in $H^2(G,M)$ provides an extension $\tilde{G}$ that induces the action on $M$; and that isn't split. Sorry Jeremy for the mixup $\endgroup$ Feb 4, 2019 at 15:51
  • $\begingroup$ @JeremyRickard yes now that I think about it this is just fine. $\endgroup$
    – hunter
    Feb 4, 2019 at 15:51

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