Timeline for Is the cross-product of two subgroups another subgroup (as claimed in the following paper)?
Current License: CC BY-SA 2.5
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 24, 2010 at 13:22 | answer | added | Jack Schmidt | timeline score: 9 | |
| Jun 24, 2010 at 11:15 | comment | added | Andrew Stacey | @Pete: I disagree! This was the question that the person was stuck on in trying to figure out something (which may well have been the finite subgroups of U(3)) and so this was the right question to ask. Working through a paper and trying to really understand everything is a very good skill to learn. The question "What are the finite subgroups of U(3)?" would mean that they probably wouldn't follow-up by working through this paper carefully. | |
| Jun 24, 2010 at 9:15 | comment | added | Pete L. Clark | P.S.: Rather than explaining an obscure sentence in a paper, maybe it will be easier for people here to answer the question: "What are the finite subgroups of $U(3)$?" Perhaps it has even been asked before... | |
| Jun 24, 2010 at 9:11 | comment | added | Pete L. Clark | I agree that the statement is incorrect or meaningless taken at face value. My best guess is that the "direct product" of two subgroups $H_1$ and $H_2$ of a group $G$ is the set $H_1 H_2$ under the conditions: $H_1 \cap H_2 = \{e\}$ and every element of $H_1$ commutes with every element of $H_2$. At least it is true that in this case $H_1 H_2$ is a subgroup which is canonically isomorphic to $H_1 \times H_2$. | |
| Jun 24, 2010 at 8:03 | comment | added | Andrew Stacey | @SCDave: but you could take a product of a subgroup with a subgroup of that subgroup. Or take a product of two subgroups that don't commute. The sentence in question (I've just looked) is a little odd as it reads as though it is a general statement that always applies, which can't be true. However, if it is to be read as only applying to the situation of the paragraph then it makes a little more sense because the subgroups in question have coprime order. (I'm not saying that that settles the matter, but that's my first impression.) | |
| Jun 24, 2010 at 7:55 | comment | added | Pietro Majer | unless it's the trivial subgroup ;) | |
| Jun 24, 2010 at 7:21 | comment | added | supercooldave | My understanding is that the definition excludes taking direct products of subgroups with themselves. Generally subgroups of a group are distinct. | |
| Jun 24, 2010 at 7:14 | history | asked | krishnamohan | CC BY-SA 2.5 |