Timeline for Maximum value of the number of conjugacy classes of nonabelian p-groups with an abelian subgroup of index p
Current License: CC BY-SA 3.0
4 events
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| Sep 19, 2011 at 19:50 | comment | added | Ralph | I'm pleased to hear that you like the answer. Thanks for your proof of $Z(G) \le A$ by contradiction: That's much better than my lengthy considerations! | |
| Sep 19, 2011 at 11:35 | comment | added | Steve D | I would just like to mention that (1) this a great answer and (2) a somewhat more general version of your theorem can be found as Lemma 4.6 in Isaacs's Finite Group Theory. The fact that $Z(G)\le A$ can also be proved by contradiction: if not, then $G=AZ(G)$ and that easily implies $G$ is abelian! | |
| Sep 18, 2011 at 21:42 | history | edited | Ralph | CC BY-SA 3.0 |
Generalized the result.
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| Sep 18, 2011 at 13:38 | history | answered | Ralph | CC BY-SA 3.0 |