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(I asked some people this question in person and got the answer "no", but wanted to see if the Internet had more to say)$ \newcommand{\FinGrp}{\mathbf{FinGrp}} $

Way back in my first group theory course, Hungerford's (undergrad) textbook explained the classification of groups as breaking up into two steps: (1) classify the finite simple groups (which has now been done), and (2) try to understand extension problems (which is also hard, and probably impossible in general).

Nowadays I know about algebraic $K$-theory, which essentially ignores step (2). Has anyone tried to understand the classification of finite simple groups through the lens of algebraic $K$-theory, or study the $K$-theory spectrum $K(\FinGrp)$? For example $K_0(\FinGrp)$ should be the free abelian group on isomorphism classes of simple groups.

There is a complication here that $\FinGrp$ is not a Waldhausen category: we have a reasonable notion of cofiber sequence $N \rightarrowtail G \twoheadrightarrow G/N$, but pushouts will not exist in general. I imagine you can fix this by working with finitely presented groups, or whatever you get from adjoining pushouts of normal-subgroup-inclusions to $\FinGrp$. Or does this tank the whole idea?

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  • $\begingroup$ Regarding the last paragraph - another possibility is to switch to the opposite category $\endgroup$
    – მამუკა ჯიბლაძე
    Commented Feb 28, 2017 at 6:22
  • $\begingroup$ "For example $K_{0}(\mathbf{FinGrp})$ should be the free abelian group on isomorphism classes of simple groups" Could you give some evidences why you believe that ? $\endgroup$
    – Ofra
    Commented Feb 28, 2017 at 20:49
  • $\begingroup$ @Ofra my group theory is a little rusty, so maybe I'm spouting nonsense. But my reasoning was as follows: if $N \rightarrowtail G \twoheadrightarrow G/N$ and $\psi$ is a generalized Euler characteristic, then $\psi(G) = \psi(N) + \psi(G/N)$, so by induction $\psi(G) = \sum_S \psi(S)$, where the sum is over the multiset of composition factors of $G$. Conversely, for a simple group $S$, we can let $\chi_S(G)$ be the number of times $S$ appears as a composition factor of $G$, and this is a well-defined Euler characteristic. I expect that $K_0(\mathbf{FinGrp})$ is generated by the $\chi_S$. $\endgroup$
    – Yuri Sulyma
    Commented Feb 28, 2017 at 20:57
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    $\begingroup$ Yes, this is not a Waldhausen category. If you add big groups, you will not get your desired $K_0$. Stick with finite groups and apply $S_\bullet$ anyhow. Try to prove you get the desired $K_0$. If all goes well, the $K$-theory of all finite groups is the $K$-theory of semi-simple groups, where all extensions split so you can use the group completion construction. That is the restricted product indexed by simple groups. It’s all about $Aut(G^n)$. For $G=Z/p$, this is $K(F_p)$. For many groups, this is $G\wr S_n$, which is somewhere in the $K$-theory literature. $\endgroup$
    – Ben Wieland
    Commented Mar 1, 2017 at 22:26

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