Prove that all groups of order $45$ are cyclic.
I was thinking about this proof for a long time and my professor said that it can be proved correctly but I’m not sure about that.
I was trying to figure it out using just:
$\gcd(45,\phi(45))$ but this is equal to $3$ not $1.$
I have read about Sylow’s theorem and I have seen someone was trying to match it with a abelian groups because $45=3^2\times 5$ etc. But I don’t know how to do it.
Let's do a description of groups of order 45. By the third Sylow's theorem, the number of subgroups of order $5$ must divide $45$, and must be congruent to $1$ modulo $5$; so there is an unique subgroup of order $5$. On the other hand, the number of subgroups of order $9$ must divide $45$ and be congruent to $1$ modulo $3$; so also there is an unique subgroup of order $9$.
The subgroup of order $5$ must be isomorphic to $\mathbb{Z}_5$ becuase that's the only group of order 5 up to isomorphism, and the subgroup of order $9$ that could be isomorphic to either $\mathbb{Z}_9$ or to $\mathbb{Z}_3\times\mathbb{Z}_3$ (since p^2 groups are abelian and the torsion coefficients of 9 could be (3,3) and (9)).
Because they are the unique p-subgroups both of them are normal; let $H_5$ be the $5$-Sylow subgroups, and $H_9$ be the $3$-Sylow subgroup. Then $H_5\cap H_9=\{1\}$ since $gcd(5,9)=1$, and $H_5H_9$ is a subgroup (both are normal). Also we know that $$|H_5H_9| = \frac{|H_5|\,|H_9|}{|{1}|} = 5\times 9 = 45 = |G|,$$ so $G=H_5H_9$. Since both are normal and have coprime orders then $G=H_5H_9\cong H_5\times H_9\cong \mathbb{Z}_5\times H_9$. Since there are two possibilities for the subgroup of order 9 then we have only two groups of order 45: $$\mathbb{Z}_5\times \mathbb{Z}_9\;\;\;\mathbb{Z}_5\times \mathbb{Z}_3 \times \mathbb{Z}_{3}$$ Both of them abelian but only the first one is cyclic, so the statement is false