Is a group of order 225 an abelian group?
I know Sylow theory and tried to use it, but the number of sylow 3 subgroups is 1 or 25.
Case $n = 3$ is 25; I can't show that G is abelian or non-abelian.
[Math] Is a group of order 225 an abelian group
abelian-groupssylow-theory
Best Answer
No, $G$ need not be abelian in general. But we can show the following. By Sylow's theorem, any group of order $225$ has a unique Sylow $5$-subgroup $P$. If this group $P$ is cyclic, then $G$ is abelian (and otherwise it need not be abelian).
Proof: 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 the order of $a$ is $25$. 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 $ab=ba$.
Now we can also show that every group of order $n$ is abelian, if and only if $n=p_1^{e_1}\cdots p_r^{e_r}$ with distinct primes $p_i$, $e_1=1,2$, and $p_i$ does not divide $p_j^{e_j}-1$ for all $i$ and all $j$. Since $225=3^25^2$, and $3\mid 5^2-1$, not every group of order $225$ is abelian. For the proof see Lemma $3.1$ in this homework, no. 4.