Timeline for Comparing the perfect groups of order 1344
Current License: CC BY-SA 4.0
11 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 17, 2023 at 18:40 | comment | added | Dave Benson | The point I was making is that Out($L$) doesn't act on $H^2(L,V)$. | |
| Jun 17, 2023 at 15:32 | comment | added | YCor | @DaveBenson but it amounts to the same when you observe that the quotient group has to be the whole automorphism group. So, you know in advance what the group is and how it acts. | |
| Jun 17, 2023 at 12:21 | comment | added | Dave Benson | @YCor Actually it's not quite like that. There are two isomorphism classes of three dimensional module for $L_3(2)$ over $\mathbb{F}_2$, but they're interchanged by the outer automorphism. So you just have to choose one and compute $H^2$, which is one dimensional. | |
| Jun 17, 2023 at 10:10 | comment | added | Derek Holt | It is hard to envisage how this problem could be solved without a massive computer calculation, which at first sight requires checking all $1344!$ possible mappings. In the end you might have to settle for an approximate answer. | |
| Jun 17, 2023 at 9:19 | comment | added | YCor | @DerekHolt Oops, thanks. So the correct conclusion is that it amounts to classify the extensions of $V=C_2^3$ by $L=\mathrm{GL}_3(\mathbf{F}_2)$ (with the standard action), so this is essentially a cohomology computation (namely of $H^2(L,V)$). If this $H^2$ has order 2 this means 2 isomorphism classes (which is what Dave says). (If it's larger, then one has to classify isomorphic groups for distinct cohomology classes, which probably amounts to check $\mathrm{Out}(L)$-orbits in this $H^2$.) | |
| Jun 17, 2023 at 8:20 | comment | added | Derek Holt | @YCor The Out of an elementary abelian group of order $8$ has order exactly $168$. | |
| Jun 17, 2023 at 8:01 | comment | added | Dave Benson | There are two isomorphism classes of perfect groups of order $1344$. They are the split and the nonsplit extensions of $L_3(2)$ by the natural module. | |
| Jun 17, 2023 at 7:38 | comment | added | YCor | The only cardinal of a nonabelian simple group dividing $1344$ is $168$, and $1344=168\times 8$. Since the Out of a group of order 8 has order $<168$, the only possibility will be a central extension. But the Schur multiplier of the simple group of order 168 has order 2. So I don't see how a perfect group of order 1344 could exist. | |
| Jun 17, 2023 at 4:05 | comment | added | Daniel Sebald | I was wondering about how similar two groups of the same “shape” really are. | |
| Jun 17, 2023 at 3:56 | comment | added | Steven Landsburg | I am dying to know how this question arose! | |
| Jun 17, 2023 at 3:26 | history | asked | Daniel Sebald | CC BY-SA 4.0 |