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 |
Questions about the branch of algebra that deals with groups.
2
votes
Is the intersection of two subgroups, defined below, always trivial?
As pointed out, the Kervaire conjecture requires an assumption about the exponent sum on $g$. So, in fact, the answer to the original question is"no".
Take $H$ be the cyclic group of order 6 generate …
1
vote
An endomorphism of free groups
Another argument:
As said above, it suffices to show that the subgroup $S=<\{g_i \alpha_i g_i^{-1}\}>$ has rank $n$. We make the subgroup graph $\Gamma_S$ of $S$ by identifying at a basepoint the roo …
-1
votes
A subgroup of corank 1 in a free group contains a primitive element?
Assuming $x_1 \notin H$, we can construct a basis for F as follows:
for each j,
if $x_j \notin H$, we set $b_j=x_j$ (in particular, $b_1=x_1$);
if $x_j \in H$, set $b_j=x_1x_j$.
Then, $B=\{b_1, …