Timeline for Order of abelian subgroup of the automorphism group of an abelian group
Current License: CC BY-SA 4.0
5 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 15 at 12:03 | comment | added | tomasz | Thanks, this is completely clear now. | |
| Mar 14 at 11:38 | comment | added | Sean Eberhard | @tomasz We have $f^i(A) \le f(p^{i-1} A) = p^{i-1} f(A) \le p^i A$ by induction. The assumption that $k \ge p$ is just used to ensure that $\binom{p^k}{i}$ is divisible by $p^k$ for $0 < i < k$. | |
| Mar 14 at 2:20 | vote | accept | tomasz | ||
| Mar 14 at 0:01 | comment | added | tomasz | Thanks, this looks very nice! Two small issues: first, you seem to be implicitly saying that $f^i$ goes to $p^iA$. Is that obvious? Secondly, I think it is worth stressing that the $1$-eigenspace of $b\in B$ cannot be all of $V$, since we are assuming that the representation is faithful. Also, how do you use the fact that $k\geq p$? | |
| Mar 13 at 14:28 | history | answered | Sean Eberhard | CC BY-SA 4.0 |