MHB Find Center of Groups of Order 8

  • Thread starter Thread starter Fessenden
  • Start date Start date
  • Tags Tags
    Center Groups
Fessenden
Messages
1
Reaction score
0
How to find the center of groups of order 8?
 
Physics news on Phys.org
Fessenden said:
How to find the center of groups of order 8?

HINT: Start by proving that if $G/Z(G)$ is cyclic then $G=Z(G)$.

This means that if $|G|=8$ then $|Z(G)|=1$, $2$ or $8$. Why can't it be $1$?
 
another way to put this is: if G is abelian, then Z(G) = G (which isn't very interesting). i believe the hint Swlabr is trying to steer you to is the class equation:

$|G| = |Z(G)| + \sum_{a \not\in Z(G)} [G:N(a)]$ where N(a) is the normalizer (or, in some texts, the centralizer) of the element a.

note each index [G:N(a)] must divide 8, and since we have the elements of Z(G) in the first term, and the a's are NOT in the center, each [G:N(a)] has to be either 2, or 4 (the only conjugacy classes of size 1 are elements of the center). in any case, each [G:N(a)] is an even number. is there an integer solution to:

8 = 1 + 2k?
 

Thread 'Trouble understanding an online solution to an exercise in Dummit & Foote'

##\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...
View full post »

Thread 'Questions about non existence of GCDs vs (coimages, cokernels)'

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...
View full post »

Thread '##(A/\mathfrak{a})_{\mathfrak{p}/\mathfrak{a}}## and its isomorphism?'

I asked online questions about Proposition 2.1.1: The answer I got is the following: I have some questions about the answer I got. When the person answering says: ##1.## Is the map ##\mathfrak{q}\mapsto \mathfrak{q} A _\mathfrak{p}## from ##A\setminus \mathfrak{p}\to A_\mathfrak{p}##? But I don't understand what the author meant for the rest of the sentence in mathematical notation: ##2.## In the next statement where the author says: How is ##A\to...
View full post »

Similar threads

A Classification of reductive groups via root datum
Replies
3
Views
2K
Insights Groups, The Path from a Simple Concept to Mysterious Results
Replies
17
Views
8K
A Center of a linear algebraic group
Replies
9
Views
2K
MHB What Is the Upper Bound of Groups of Order in Finite Group Theory?
Replies
2
Views
2K
I Dfns group action & group, written in terms of the other
Replies
26
Views
377
A Near-Rings with Noncommutative Addition and Two-Sided Distributivity
Replies
4
Views
2K
MHB Is Every Group of Order 25 Cyclic?
Replies
1
Views
1K
MHB How to Determine the Order of \( g^8 \) in a Group?
Replies
3
Views
2K
I Isomorphisms between C4 & Z4 Groups
Replies
1
Views
2K
I Is There a Connection Between Conjugation and Change of Basis?
Replies
1
Views
3K

Hot Threads

Recent Insights

Back
Top