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Thread: Algebra, Problems For Fun (22)

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

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


    Source: MathLinks


    Suggestion:

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

    $\displaystyle x^9 \equiv 1 \mod 25$ 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 $\displaystyle |G|=5^{2} \cdot 3^{2}$. Clearly it has unique 5-sylow subgroup and hence its normal in $\displaystyle G.$ 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 $\displaystyle P \cap Q=(e)$ and since they are normal so $\displaystyle pqp^{-1}q^{-1}=(e)$.

    Now if the number of 3-sylow subgroups are 25 then by a lemma given in hersteins "topics in algebra book" we see that $\displaystyle G$ cannot be simple as $\displaystyle |G| \nmid i(H)!$. 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 $\displaystyle |G|=5^{2} \cdot 3^{2}$. Clearly it has unique 5-sylow subgroup and hence its normal in $\displaystyle G.$ 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 $\displaystyle P \cap Q=(e)$ and since they are normal so $\displaystyle pqp^{-1}q^{-1}=(e)$.

    Now if the number of 3-sylow subgroups are 25 then by a lemma given in hersteins "topics in algebra book" we see that $\displaystyle G$ cannot be simple as $\displaystyle |G| \nmid i(H)!$. But i am unable to deduce this part.
    $\displaystyle G$ 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 $\displaystyle Q.$ let $\displaystyle P=<a>$ be the Sylow 5 subgroup. so $\displaystyle G=PQ$ and both $\displaystyle P, Q$ are abelian. in order to prove that $\displaystyle G$ is abelian,

    we only need to prove that $\displaystyle ab=ba,$ for any $\displaystyle b \in Q.$ now $\displaystyle bab^{-1}=a^k,$ for some $\displaystyle k,$ because $\displaystyle P$ is normal in $\displaystyle G.$ also $\displaystyle b^9=1,$ because $\displaystyle |Q|=9$ and $\displaystyle b \in Q.$ thus $\displaystyle a=b^9ab^{-9}=a^{k^9}$ and so $\displaystyle a^{k^9 - 1} = 1.$

    hence $\displaystyle k^9 \equiv 1 \mod 25,$ because $\displaystyle o(a)=25.$ this is now an easy number theory problem to see that $\displaystyle k^9 \equiv 1 \mod 25$ has only one solution: $\displaystyle k \equiv 1 \mod 25.$ therefore $\displaystyle bab^{-1}=a^k=a. \ \ \Box$
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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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