Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Normalizer of a subgroup

  1. Nov 30, 2011 #1
    1. The problem statement, all variables and given/known data

    Prove the number of distinct conjugate subgroups of a subgroup H in a group G is [G:N(H)] where N(H)={g [itex]\in[/itex] G | gHg[itex]^{-1}[/itex]=H}.

    2. Relevant equations

    I'm thinking the counting formula; G=|C(x)||Z(x)| with C(x) being the conjugacy class of x and Z(x) being the centralizer of x.

    3. The attempt at a solution

    I thought that I could say N(H) is the centralizer of H and [G:N(H)]=|G|/|N(H)| so by the counting formula |G|/|N(H)|=|C(H)| and |C(H)| is the number of distinct conjugate subgroups H. In truth I have only worked with the counting formula for elements, never with subgroups. Also if this does hold somewhat true, it seems to me to only apply to the case when G is finite. What about when G is infinite?
  2. jcsd
  3. Dec 2, 2011 #2
    Should I post this in the Linear Algebra thread? The rules say not to but it seems to me that questions such as this get answered there. I'm really just looking for if my solution is valid or not as well as a hint for the infinite case..
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook