Group of order 45 is abelian

Question

I am trying to prove that the inverse of Lagrange Theorem holds for a group $|G|$ of order $45$ and that it contains a subgroup of order $3$. I think I proved that $G$ is abelian and therefore the inverse of Lagrange holds but I am not so sure about my proof. So, since $45=5\times 3^{2}$ we can easily prove that there is a unique 5-Sylow subgroup of order 5 (let's call this $H$) and a unique 3-Sylow subgroup of order 9 (let's call this $K$), both of which are normal. (They are also abelian since $|H|=5$, prime and thus cyclic and $|K|=9=3^{2}$ and we know that every group of order $p^{2}$ is abelian.) Now since they are normal $HK$ is a subgroup of $G$ and since $5$, $9$ are coprime we have that $H\cap K= \{1_{G} \}$. Thus, since $$|HK|=\frac{|H||K|}{|H\cap K|} $$ we have that $HK=G$. Now we prove that $HK$ is abelian. Firstly, we have $hk=kh$ $\forall h\in H, k\in K$. Indeed we have that $(h,k)=hkh^{1}k^{-1} \in H$ and $ (h,k)\in K$ beacause $H, K$ are normal and thus $(h,k)\in H\cap K =\{1_{G}\}$. Now, if $g_{1}=h_{1}k_{1}, g_{2}=h_{2}k_{2} \in G$ then $$g_{1}g_{2}=h_{1}k_{1}h_{2}k_{2}=h_{2}k_{2}h_{1}k_{1}=g_{2}g_{1}$$ using the fact that $H,K $ are abelian and that $hk=kh$. Therefore $G$ is abelian. Then to prove that there is a subgroup of order $3$ we find the non isomorphic abelian groups of order $45$ which are $G_{1}=\mathbb{Z_{5}}\times\mathbb{Z_{9}}\cong\mathbb{Z_{45}}$ and $G_{2}=\mathbb{Z_{5}}\times\mathbb{Z_{3}}\times\mathbb{Z_{3}}$ which both have $\mathbb{Z_{3}}$ of order $3$ as a subgroup. Is my proof correct? Thank you in advance for your time.

Answer 1

Your proof certainly does work. However, you can do it without using the classification of abelian groups of order $45$. We need to find subgroups of order $3$, $5$, $9$, and $15$. A comment showed you how to do order $3$; you can also use Cauchy's theorem. $5$ and $9$ are covered by Sylow's theorems, and you can use the fact that the subgroup of order $5$ is normal and your formula for the order of the subgroup generated by the subgroups of order $3$ and $5$ to get one of order $15$.

Answer 2

As you have proved already that $H$ and $K$ are normal, you just need to recall the lemmas stating that a) normal subgroups are union of conjugacy classes, and b) from $G/Z(G)$ cyclic follows $G$ is abelian. In fact, by contradiction suppose $G$ is non-abelian. Then:

• $|Z(G)|=1$ is ruled out because there aren't conjugacy classes of size $2$ or $4$, and hence $H$ (of order $5=1+4$) can't be built up as union of conjugacy classes;
• $|Z(G)|=3$ is ruled out by the argument above, considered that $H\cap Z(G)=\{1\}$;
• $|Z(G)|=5$ is ruled out by the same argument above, considered that $K\cap Z(G)=\{1\}$, and there aren't conjugacy classes of size $8$, either;
• $|Z(G)|=3^2$ is ruled out because $G/Z(G)$ would be cyclic, and hence $G$ abelian (a contradiction in this assumption);
• $|Z(G)|=3.5$ is ruled out by the argument in the previous point.

So, $Z(G)=G$.