Thread: Algebra, Problems For Fun (22)

1. 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.

Suggestion:

Spoiler:
Let $P=.$ 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.

Note: Obviously you'll have time to post your solution as long as it has not been answered in MathLinks!

2. 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.

3. Originally Posted by Chandru1
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 $P=$ 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 $bab^{-1}=a^k=a. \ \ \Box$

4. Ok

Hi--

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