Search Results
Search type | Search syntax |
---|---|
Tags | [tag] |
Exact | "words here" |
Author |
user:1234 user:me (yours) |
Score |
score:3 (3+) score:0 (none) |
Answers |
answers:3 (3+) answers:0 (none) isaccepted:yes hasaccepted:no inquestion:1234 |
Views | views:250 |
Code | code:"if (foo != bar)" |
Sections |
title:apples body:"apples oranges" |
URL | url:"*.example.com" |
Saves | in:saves |
Status |
closed:yes duplicate:no migrated:no wiki:no |
Types |
is:question is:answer |
Exclude |
-[tag] -apples |
For more details on advanced search visit our help page |
For questions about groups whose elements commute.
0
votes
Maximal subgroups of a finite p-group
A s noted Prof. Robinson, this is false. However, this is true iff $L\not\le M\Phi(G)$. Indeed, let $\bar G=G/M\Phi(G)$; then $\bar L$ is a direct factor of $\bar G$ of order $p$. If $\bar G=\bar L\ti …
1
vote
Cyclic subgroups of finite abelian groups
If $G$ is an abelian $p$-group and $p^k\le\exp(G)$, then the number of cyclic subgropups of order $p^k$ in $G$ is
$$
{\rm c}_k(G)=\frac{|\Omega_k(G)-\Omega_{k-1}(G)|}{(p-1)p^{k-1}}.
$$
If the type of …
3
votes
Is there a nice explanation for this curious fact about cyclic subgroups?
Another proof of Strickland's result.
Let $G$ be a group of order $p^n$ and $\nu(G)$ be the sum of orders of its cyclic subgroups. To prove that $\nu(G)=\sigma_1(G)$, we proceed by induction on $|G|$ …