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Questions about the branch of algebra that deals with groups.
3
votes
1
answer
258
views
Nonisomorphic central products on the same pair of groups?
A central product of two groups $G$ and $H$ is determined as follows. The groups $G$ and $H$ have respective central subgroups $A$ and $B$ which are isomorphic, let $\delta:A\rightarrow B$ be such a …
6
votes
2
answers
340
views
Differences between $p$-groups and $q$-groups
First, let me include the same disclaimer that goes in the first line of any article I write: all groups considered herein are finite.
Academically, I work with connecting the arithmetic structure of …
4
votes
Is a point stabilizer in the Mathieu group $M_{20}$ half-transitive?
I certainly appreciate Max's answer (especially the geometric interpretation) and have accepted it as such. However, as is so usual, writing up the question caused me to focus on it with greater clar …
4
votes
3
answers
326
views
Is a point stabilizer in the Mathieu group $M_{20}$ half-transitive?
The background: We recall/define the following:
$\Omega_n=\{1,\dots,n\}$.
$M_n$ is the Mathieu group of degree $n$. We follow the Wikipedia article "Mathieu group" and define these groups for each …
3
votes
Cancellation theorem for direct and other kinds of products between groups
For a straightforward example that the subgroup doing the acting in a semidirect product cannot automatically be cancelled (and as semidirect products are knit products, also addresses that case), the …
4
votes
0
answers
267
views
Has the external knit product been used to construct a previously unknown group?
In the Wikipedia article
Zappa–Szép product
, the knit product (a.k.a. Zappa–Szép product, Zappa–Rédei-Szép product, general product, exact factorization) is defined, and its basic properties are laid …
31
votes
2
answers
1k
views
Does the symmetric group $S_{10}$ factor as a knit product of symmetric subgroups $S_6$ and ...
By knit product (alias: Zappa-Szép product), I mean a product $AB$ of subgroups for which $A\cap B=1$. In particular, note that neither subgroup is required to be normal, thus making this a generaliz …