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  1. #1
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    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)
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  2. #2
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    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 a^{-1}ba=b^i \neq b. show that i=2 works. then the group generated by a,b with those relations would be a (actually "the") non-abelian group of order 21.
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  3. #3
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    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, P \in Syl_3(G) , and let 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 G = P \times Q. However, (q-1)=6 is divisible by p=3.

    In this case, we use a semidirect product and G = Q \rtimes_{\varphi} P \cong C_7 \rtimes_{\varphi} C_3, for some \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 Aut(C_7) \cong \mathbb{Z}_6 and  |Aut(C_7)| is divisible by |\phi(x)| , where x in C_3, we see that \phi is not trivial.

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

    Consider ba = (1,x)(y,1) = (1 \phi(x)(y), x ) = (y^2, x) = a^2b. We see that \phi:C_3 \rightarrow Aut(C_7) is defined by  x \mapsto (y \mapsto y^2) here since 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 G=<a, b |a^7=1, b^3=1, bab^{-1} = a^2>, where |G|=21.
    Last edited by aliceinwonderland; October 15th 2009 at 09:49 PM. Reason: addition
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