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

Posted: December 13, 2010 in Elementary Algebra; Problems & Solutions, Groups and Fields
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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 $ab=ba. \ \Box$

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?

• Yaghoub says:

I didn’t say that! I said $bab^{-1}=a^k$ and that’s because $P$ is normal.