Groups of order 225 with cyclic Sylow 5-subgroup are abelain

Posted: December 13, 2010 in Elementary Algebra; Problems & Solutions, Groups and Fields
Tags: , ,

Problem. Let G be a group of order 225. By Sylow theorem G has a unique Sylow 5-subgroup P. Prove that if P is cyclic, then G is abelian.

Solution. Choose any Sylow 3-subgroup Q and let P = \langle a \rangle. So G=PQ and both P and Q are abelian. In order to prove that G is abelian, we only need to prove that ab=ba, for any b \in Q. Now bab^{-1}=a^k, for some k, because P is normal in G. Also b^9=1 because |Q|=9. Thus a=b^9ab^{-9}=a^{k^9} and so a^{k^9 - 1} = 1. Hence k^9 \equiv 1 \mod 25, because o(a)=25. It is easy to see that k^9 \equiv 1 \mod 25 has only one solution, i.e. k \equiv 1 \mod 25. Therefore bab^{-1}=a^k=a and so ab=ba. \ \Box

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Comments
  1. zeynep says:

    i didn’t get that part (b^9)a(b^-9)=(a)^(k^9)

  2. Anjit kumar jha says:

    I have not ideas about sylow 5- subgroup and sylow 3 -subgroup of a group of order 225.

  3. peter says:

    Why b^9ab^-1=a^k?

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