MHB Is the Action of Conjugation by Sylow 2-Subgroups Onto?

  • Thread starter Thread starter Poirot1
  • Start date Start date
  • Tags Tags
    Group Image
Poirot1
Messages
243
Reaction score
0
Assume that G of order 48 has 3 sylow 2-subgroups. Let G act on the set of such subgroups by conjugation. How do I know that this action is onto? I know that all 3 subgroups are conjugate but I'm not sure this is enough.
 
Physics news on Phys.org
I can't imagine this is too difficultto find that an action is onto i.e. can I map to every permutation in S3. My problem is (1 2 3) is in S3 so is this telling me I need to find a g such that gHg^-1=M, gMg^-1=K and gkg^-1=H, for the slylow subgroups H,k,M
 
ok, suppose the three subgroups are H,K,M.

we know there is x,y,z in G with:

xHx-1 = K

yKy-1= M

zMz-1= H.

so the action is transitive. because we have an action (these are bijective maps), if:

x:H-->K

then xKx-1 can only be H or M. if xKx-1 = H, then x corresponds to the permutation (1 2).

otherwise xKx-1 = M, and x corresponds to (1 2 3).

now if xKx-1 corresponds to (1 2), we can look at y.

since y takes K to M, either y takes H to K, in which case y corresponds to (1 2 3)

or y takes H to H, in which case y corresponds to (2 3).

in this last case, we have xy corresponds to (1 2)(2 3) = (1 2 3):

xyH(xy)-1 = x(yHy-1)x-1 = xHx-1 = K

xyK(xy)-1 = x(yKy-1)x-1 = xMx-1 = M

xyM(xy)-1 = x(yMy-1)x-1 = xKx-1 = H.

(and we didn't even need to use z).
 
##\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