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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


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    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


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    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


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    Of course.
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