Results 1 to 3 of 3

Thread: nonabelian group

  1. #1
    Junior Member
    Joined
    Sep 2009
    Posts
    34

    nonabelian group

    Construct a nonabelian group of order 21 ( Hint : Assume that a^3 = e, b^7 = e and find some i such that a^-1ba =a^i not equal a, which is consistent with the relation a^3 = a^7 = e)
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    May 2008
    Posts
    2,295
    Thanks
    7
    Quote Originally Posted by Godisgood View Post

    Construct a nonabelian group of order 21 ( Hint : Assume that a^3 = e, b^7 = e and find some i such that a^-1ba =a^i not equal a, which is consistent with the relation a^3 = a^7 = e)
    that part in red is wrong. it should be $\displaystyle a^{-1}ba=b^i \neq b.$ show that $\displaystyle i=2$ works. then the group generated by $\displaystyle a,b$ with those relations would be a (actually "the") non-abelian group of order 21.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Senior Member
    Joined
    Nov 2008
    Posts
    394
    Quote Originally Posted by Godisgood View Post
    Construct a nonabelian group of order 21
    There is a systematic mechanism when classifying groups of order pq, p and q primes with p < q and p | q-1.

    Below is to find a nonabelian group of order 21 using a semidirect product.

    Let G be group of order 21. Let P be a Sylow-3 subgroup of G, $\displaystyle P \in Syl_3(G) $, and let $\displaystyle Q \in Syl_7(G) $. If (q-1)=6 were not divisible by p=3. It is easy. P and Q are unique sylow subgroups in G and $\displaystyle G = P \times Q$. However, (q-1)=6 is divisible by p=3.

    In this case, we use a semidirect product and $\displaystyle G = Q \rtimes_{\varphi} P \cong C_7 \rtimes_{\varphi} C_3$, for some $\displaystyle \phi:C_3 \rightarrow Aut(C_7)$ (We know that Q is a unique sylow-7 subgroup in G, thus it is normal in G. However, P is not a unique sylow-3 subgroup in G).
    Since $\displaystyle Aut(C_7) \cong \mathbb{Z}_6$ and $\displaystyle |Aut(C_7)| $ is divisible by $\displaystyle |\phi(x)| $, where x in $\displaystyle C_3$, we see that $\displaystyle \phi$ is not trivial.

    Let $\displaystyle C_3 = <x>$, $\displaystyle C_7=<y>$; let $\displaystyle a=(y,1), b=(1,x)$. Then $\displaystyle G=<a,b>$ with some relationships, where $\displaystyle a^7= 1 $ and $\displaystyle b^3=1$

    Consider $\displaystyle ba = (1,x)(y,1) = (1 \phi(x)(y), x ) = (y^2, x) = a^2b$. We see that $\displaystyle \phi:C_3 \rightarrow Aut(C_7)$ is defined by $\displaystyle x \mapsto (y \mapsto y^2)$ here since $\displaystyle Aut(C_7)$ is cyclic, containing a unique subgroup of order 3 by Cauchy's theorem; the latter parenthesis is used to denote the automorphism of order 3.

    Thus we have $\displaystyle G=<a, b |a^7=1, b^3=1, bab^{-1} = a^2>$, where $\displaystyle |G|=21$.
    Last edited by aliceinwonderland; Oct 15th 2009 at 09:49 PM. Reason: addition
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. [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
  2. Replies: 1
    Last Post: Nov 4th 2009, 09:52 AM
  3. Nonabelian group of order p^3
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: Nov 19th 2008, 05:17 AM
  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 for this page

Search Tags


/mathhelpforum @mathhelpforum