1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

P-group problem group theory

  1. Jan 31, 2015 #1
    1. The problem statement, all variables and given/known data
    Let G be a non-abelian group with order ##p^3##, p prime. Then show that the order of the center must be p.

    2. Relevant equations
    Theorem in our book says that for any p-group the center is non-trivial and it's order is divisible by p.
    Class eq.
    ##|G|= |Z(G)| + \sum{[G : C(x)]}## Where, [G: C(x)] is the number of left cosets of C(x) in G. And ##C(x) := (a\in G | axa^{-1}=x)## is the centralizer. And we are summing over x by taking a single x in each of the non-trivial conjugacy class of G and looking at that centralizer.

    3. The attempt at a solution
    My thought was to do a counting argument, but it felt really weak.

    So by the theorem in the book, since G is a p-group we have that Z(G) is non trivial and p | |Z(G)|, and obviously the order of Z(G) is strictly less that G since G is non-abelian. So we get that |Z(G)| must be either ##p## or ##p^2##.

    So, for contradiction, suppose that ##|Z(G)|=p^2##. So we have $$p^3= p^2+ \sum{[G : C(x)]}$$ $$p^3-p^2=\sum{[G : C(x)]}$$
    From here I apply Lagrange's theorem ##[G:C(x)]=\frac{|G|}{|C(x)|}## and reason that since any centralizer has at the very least, the center of the group, $$\frac{|G|}{|C(x)|}<\frac{p^3}{p^2}=p$$. I further reason that there are ##p^3-p^2## non-central elements, and since a non-trivial conjugacy class has at least 2 elements in it. That there are at most ##\frac{p^3-p^2}{2}## non-trivial conjugacy classes. So the sum$$\sum{[G : C(x)]}<\frac{p^3-p^2}{2}p$$ Giving the inequality:$$p^3-p^2<\frac{p^3-p^2}{2}p$$. Which fails when p=2. Thus a contradiction.

    The reason this feels weak, is because it doesnt fail for p=3, or p=5, or any other primes I tried.

    A classmate mentioned an argument using the fact that a group mod its center is abelian, but I'm not sure that is true in general. I couldn't find it anywhere.
     
  2. jcsd
  3. Jan 31, 2015 #2

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Think about the factor group ##G/Z(G)## when ##|Z(G)|=p^2##. A group mod its center isn't always abelian. But this one is. In fact, it's cyclic. What can you do with that?
     
    Last edited: Jan 31, 2015
  4. Jan 31, 2015 #3
    Well, if G/Z(G) is cyclic, then its trivial, meaning that G is abelian, which is a contradiction.

    How do we know that G/Z(G) is cyclic in this case?
     
  5. Jan 31, 2015 #4

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    What's the order of ##G/Z(G)##?
     
  6. Feb 10, 2015 #5
    Sorry for the delay in response. In this case, G/Z(G) had order p-prime. Thus, it is cyclic.
     
  7. Feb 10, 2015 #6

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Of course.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: P-group problem group theory
  1. Group theory problem (Replies: 0)

Loading...