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Math Help - Algebra, Problems For Fun (22)

  1. #1
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    Algebra, Problems For Fun (22)

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


    Source: MathLinks


    Suggestion:

    Spoiler:
    Let P=<a>. Let Q be a Sylow 3 subgroup of G. Then PQ=G. For any b \in Q we have bab^{-1}=a^k, for some k. now use this fact that x \equiv 1 \mod 25 is the unique solution of

    x^9 \equiv 1 \mod 25 to finish the proof.


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  2. #2
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    Solution

    Ok. This is what i think the solution should be an d i think i have made a mistake in the last part. Just see whats wrong and recitfy.

    We have |G|=5^{2} \cdot 3^{2}. Clearly it has unique 5-sylow subgroup and hence its normal in G. Now if we work out the number of 3 sylow subgroups one can see that its either 1 or 25 such subgroups are possible. If its one then your question is trivial since P \cap Q=(e) and since they are normal so pqp^{-1}q^{-1}=(e).

    Now if the number of 3-sylow subgroups are 25 then by a lemma given in hersteins "topics in algebra book" we see that G cannot be simple as |G| \nmid i(H)!. But i am unable to deduce this part.
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  3. #3
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    Quote Originally Posted by Chandru1 View Post
    Ok. This is what i think the solution should be an d i think i have made a mistake in the last part. Just see whats wrong and recitfy.

    We have |G|=5^{2} \cdot 3^{2}. Clearly it has unique 5-sylow subgroup and hence its normal in G. Now if we work out the number of 3 sylow subgroups one can see that its either 1 or 25 such subgroups are possible. If its one then your question is trivial since P \cap Q=(e) and since they are normal so pqp^{-1}q^{-1}=(e).

    Now if the number of 3-sylow subgroups are 25 then by a lemma given in hersteins "topics in algebra book" we see that G cannot be simple as |G| \nmid i(H)!. But i am unable to deduce this part.
    G is not simple because Sylow 5-subgroup is normal. if you're trying to prove that Sylow 3-subgroup is normal too, then i don't think that would be very easy! if you follow my suggestion, then

    you'll see the problem is not that hard: choose any Sylow 3-subgroup Q. let P=<a> be the Sylow 5 subgroup. so G=PQ and both P, 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 and b \in Q. 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. this is now an easy number theory problem to see that k^9 \equiv 1 \mod 25 has only one solution: k \equiv 1 \mod 25. therefore bab^{-1}=a^k=a. \ \ \Box
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  4. #4
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    Ok

    Hi--

    Yes! actually i had that in mind but wasnt able to put forward! Sorry for the inconvinience
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