Timeline for Is every finite group the outer automorphism group of a finite group?

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May 8 at 17:41	history	edited	LSpice	CC BY-SA 4.0	Links, while this is on the front page
Sep 24, 2020 at 15:27	comment	added	Nicholas Kuhn		@YCor Thanks for pointing out this classic fact. I've added this into my answer.
Sep 24, 2020 at 15:26	history	edited	Nicholas Kuhn	CC BY-SA 4.0	added 137 characters in body
Sep 24, 2020 at 15:07	comment	added	YCor		Also the induced map $\mathrm{Out}(P)\to\mathrm{Aut}(P/\Phi P)$ has its kernel a $p$-group. Hence this shows that for every finite group $G$ and prime $p$ there exists a finite $p$-group $H$ and a surjective map $\mathrm{Out}(H)\to G$ whose kernel is a $p$-group.
Sep 24, 2020 at 15:04	history	edited	Nicholas Kuhn	CC BY-SA 4.0	added 286 characters in body
Sep 24, 2020 at 15:02	comment	added	Nicholas Kuhn		@Carl-FredrikNybergBrodda Thanks for looking this up. So I guess they basically say the same as Bryant and Kovacs.
Sep 24, 2020 at 14:59	history	edited	Nicholas Kuhn	CC BY-SA 4.0	added 286 characters in body
Sep 24, 2020 at 13:52	comment	added	Carl-Fredrik Nyberg Brodda		(The book is available online, albeit from canonical non-MO-linkable places).
Sep 24, 2020 at 13:50	comment	added	Carl-Fredrik Nyberg Brodda		A quote from the remark at the end of §13 of the above reference: "Theorem 13.5 shows that any subgroup of $GL(n, p)$ is the linear group induced on $\mathfrak{P}/ \Phi(\mathfrak{P})$ by Aut$(\mathfrak{P})$ for some $p$-group $\mathfrak{P}$. Since any finite group is isomorphic to a subgroup of $GL(n, p)$ for some $n$, every finite group is isomorphic to the group induced on $\mathfrak{P} / \Phi(\mathfrak{P})$ by Aut$(\mathfrak{P})$ for some $p$-group $\mathfrak{P}$."
Sep 24, 2020 at 13:36	history	answered	Nicholas Kuhn	CC BY-SA 4.0