Results 1 to 2 of 2

Thread: Nonabelian group of order p^3

  1. #1
    Senior Member
    Joined
    Nov 2008
    Posts
    394

    Nonabelian group of order p^3

    [Hungerford, p99 Q7]

    Show that

    If G is a nonabelian group of order $\displaystyle p^{3}$ (p prime), then the center of G is the subgroup generated by all elements of the form $\displaystyle aba^{-1}b^{-1} (a,b \in G)$.
    Last edited by aliceinwonderland; Nov 18th 2008 at 08:10 PM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    10
    Quote Originally Posted by aliceinwonderland View Post
    If G is a nonabelian group of order $\displaystyle p^{3}$ (p prime), then the center of G is the subgroup generated by all elements of the form $\displaystyle aba^{-1}b^{-1} (a,b \in G)$.
    Let $\displaystyle C$ be the subgroup generated by elements of the form $\displaystyle aba^{-1}b^{-1}$ i.e. commutator subgroup. By Burnside's lemma $\displaystyle Z(G)$, the center, is non-trivial. Therefore, $\displaystyle |Z(G)|\geq p$. But then the group $\displaystyle |G/Z(G)|$ has order $\displaystyle p\text{ or }p^2$ and therefore it is abelian. Thus, we must have that $\displaystyle C\subseteq Z(G)$. We are told that $\displaystyle G$ is not abelian therefore, $\displaystyle C\not = \{ 1 \}$ and $\displaystyle Z(G) \not = G$. Suppose that $\displaystyle |Z(G)|=p^2$ and choose $\displaystyle a\in G - Z(G)$. Then $\displaystyle C(a)$, the centralizer, contains $\displaystyle Z(G)$ properly, and is a subgroup of $\displaystyle G$. Thus, Lagrange's theorem forces $\displaystyle |C(a)| = p^3$. But that would mean the index of $\displaystyle C(a)$ in $\displaystyle G$ is $\displaystyle 1$. This means that the conjugacy class of $\displaystyle a$ is just $\displaystyle \{ a \}$ and so $\displaystyle a\in Z(G)$ - a contradiction! Thus, we see that $\displaystyle |Z(G)| \not = p^2$. The only remaining possibility is that $\displaystyle |Z(G)| = p$. However, $\displaystyle C$ is a non-trivial subgroup of $\displaystyle Z(G)$ which forces $\displaystyle C = Z(G)$.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Order of Group. Direct Product of Cyclic Group
    Posted in the Advanced Algebra Forum
    Replies: 9
    Last Post: Nov 19th 2011, 01:06 PM
  2. [SOLVED] Show that a nonabelian group must have at least five distinct elements.
    Posted in the Advanced Algebra Forum
    Replies: 4
    Last Post: Nov 11th 2010, 01:32 PM
  3. nonabelian group
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: Oct 15th 2009, 08:26 PM
  4. Nonabelian group of order p^3 (2)
    Posted in the Advanced Algebra Forum
    Replies: 0
    Last Post: Nov 18th 2008, 07:49 PM
  5. Nonabelian group of order 12
    Posted in the Advanced Algebra Forum
    Replies: 0
    Last Post: Nov 18th 2008, 07:40 PM

Search Tags


/mathhelpforum @mathhelpforum