Is direct product of subgroups a subgroup?

krishna mohan
Messages
114
Reaction score
0
Hi..

In the second paragraph of the following paper, there is a statement: "Because the direct product of subgroups is automatically a subgroup.."

http://jmp.aip.org/jmapaq/v23/i10/p1747_s1?bypassSSO=1

I don't see how that can be true...you can always take direct product of a subgroup with itself many times and create a group of order larger than the parent group...
 
Last edited by a moderator:
Physics news on Phys.org
If H_1 is a subgroup of G_1 and H_2 is a subgroup of G_2 then H_1 x H_2 is a subgroup of G_1 x G_2. In particular, if H1 and H2 are subgroups of a group G, then H1 x H2 is a subgroup of G x G (you are right, not of G itself).

I haven't read the article, but the authors might have meant this.
 
##\textbf{Exercise 10}:## I came across the following solution online: Questions: 1. When the author states in "that ring (not sure if he is referring to ##R## or ##R/\mathfrak{p}##, but I am guessing the later) ##x_n x_{n+1}=0## for all odd $n$ and ##x_{n+1}## is invertible, so that ##x_n=0##" 2. How does ##x_nx_{n+1}=0## implies that ##x_{n+1}## is invertible and ##x_n=0##. I mean if the quotient ring ##R/\mathfrak{p}## is an integral domain, and ##x_{n+1}## is invertible then...
The following are taken from the two sources, 1) from this online page and the book An Introduction to Module Theory by: Ibrahim Assem, Flavio U. Coelho. In the Abelian Categories chapter in the module theory text on page 157, right after presenting IV.2.21 Definition, the authors states "Image and coimage may or may not exist, but if they do, then they are unique up to isomorphism (because so are kernels and cokernels). Also in the reference url page above, the authors present two...
When decomposing a representation ##\rho## of a finite group ##G## into irreducible representations, we can find the number of times the representation contains a particular irrep ##\rho_0## through the character inner product $$ \langle \chi, \chi_0\rangle = \frac{1}{|G|} \sum_{g\in G} \chi(g) \chi_0(g)^*$$ where ##\chi## and ##\chi_0## are the characters of ##\rho## and ##\rho_0##, respectively. Since all group elements in the same conjugacy class have the same characters, this may be...
Back
Top